4

The following code from this example (Plane partition by Jang Soo Kim) has 2 faults, I could not figure out how to overcome them.

Firsty, the value 0 equals to value 2; i.e. value 0 draws 2 cubes.

Secondly, it does not allow for drawing a void cube position, i.e. to generate the second drawing.

\documentclass{beamer}
\setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
\usepackage{tikz}
\usepackage{verbatim}
% Three counters
\newcounter{x}
\newcounter{y}
\newcounter{z}
% The angles of x,y,z-axes
\newcommand\xaxis{210}
\newcommand\yaxis{-30}
\newcommand\zaxis{90}
% The top side of a cube
\newcommand\topside[3]{
  \fill[fill=yellow, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
\newcommand\leftside[3]{
  \fill[fill=green, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
\newcommand\rightside[3]{
  \fill[fill=blue, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube 
\newcommand\cube[3]{
  \topside{#1}{#2}{#3} \leftside{#1}{#2}{#3} \rightside{#1}{#2}{#3}
}
% Definition of \planepartition
% To draw the following plane partition, just write \planepartition{ {a, b, c}, {d,e} }.
%  a b c
%  d e
\newcommand\planepartition[1]{
 \setcounter{x}{-1}
  \foreach \a in {#1} {
    \addtocounter{x}{1}
    \setcounter{y}{-1}
    \foreach \b in \a {
      \addtocounter{y}{1}
      \setcounter{z}{-1}
      \foreach \c in {1,...,\b} {
        \addtocounter{z}{1}
        \cube{\value{x}}{\value{y}}{\value{z}}
      }
    }
  }
}
\begin{document} 
\begin{tikzpicture}
\planepartition{{2,1,2,1,2,1,2},{1,1,1,1,1,1,1},{1,1,1,1,1,1,1},{1,1,1,1,1,1,1},{2,1,2,1,2,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
\end{tikzpicture}
\end{document} 

enter image description here enter image description here

AndréC
  • 24,137
Hany
  • 4,709

1 Answers1

3

Please note that there is a follow-up question here...

There's only one mistake (and not 2 faults) in the \foreach loop. Indeed when it enters the third loop with \b=0

\foreach \c in {1,..., \b} 

the foreach becomes \c in {1,0} which corresponds to 2 iterations.

All you have to do is place a test to prevent iterations when \b = 0.

\newcommand\planepartition[1]{
 \setcounter{x}{-1}
  \foreach \a in {#1} {
        \addtocounter{x}{1}
        \setcounter{y}{-1}
        \foreach \b in \a {
            \addtocounter{y}{1}
            \setcounter{z}{-1}
            \ifnum \b>0
            \foreach \c in {1,...,\b} {
                \addtocounter{z}{1}
                \cube{\value{x}}{\value{y}}{\value{z}}
      }\fi
    }
  }
}

cubes

\documentclass{beamer}
\setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
\usepackage{tikz}
\usepackage{verbatim}
% Three counters
\newcounter{x}
\newcounter{y}
\newcounter{z}
% The angles of x,y,z-axes
\newcommand\xaxis{210}
\newcommand\yaxis{-30}
\newcommand\zaxis{90}
% The top side of a cube
\newcommand\topside[3]{
  \fill[fill=yellow, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
\newcommand\leftside[3]{
  \fill[fill=green, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
\newcommand\rightside[3]{
  \fill[fill=blue, draw=black,shift={(\xaxis:#1)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube 
\newcommand\cube[3]{
  \topside{#1}{#2}{#3} \leftside{#1}{#2}{#3} \rightside{#1}{#2}{#3}
}
% Definition of \planepartition
% To draw the following plane partition, just write \planepartition{ {a, b, c}, {d,e} }.
%  a b c
%  d e
\newcommand\planepartition[1]{
 \setcounter{x}{-1}
  \foreach \a in {#1} {
        \addtocounter{x}{1}
        \setcounter{y}{-1}
        \foreach \b in \a {
            \addtocounter{y}{1}
            \setcounter{z}{-1}
            \ifnum \b>0
            \foreach \c in {1,...,\b} {
                \addtocounter{z}{1}
                \cube{\value{x}}{\value{y}}{\value{z}}
      }\fi
    }
  }
}
\begin{document} 
\begin{tikzpicture}
\planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
\end{tikzpicture}
\end{document} 
AndréC
  • 24,137
  • 1
    Thank you very much for your answer. I also found that adding , opacity=.7 in the \fill options gives a good effect. – Hany Jan 20 '19 at 06:15
  • This creates a transparent effect as if the sides were made of coloured glass. That's nice. – AndréC Jan 20 '19 at 06:18
  • Please refer to my question https://tex.stackexchange.com/questions/470983/modify-the-code-of-the-example-plane-partition-by-jang-soo-kim-from-texample-ne – Hany Jan 20 '19 at 06:43