Replicating this Visual Proof with TikZ

Question

I'm fairly new to TikZ and would like to replicate the proof without words shown here.

Is there any way to construct these clean cubes in 3D without drawing them line by line?

Thank you for your help with this :)

Welcome to TeX.SX! What have your tried so far? Note that on this site do-it-for-me questions that do not show some own effort are discouraged. — siracusa, Jul 12 '19 at 18:26
This is possible with the construction of Jang Soo Kim whose code has been fixed here: https://tex.stackexchange.com/q/470653/138900 — AndréC, Jul 12 '19 at 18:57
As well as this one which allows to place the cubes where you want, i.e. with holes or no cube underneath like on this question: https://tex.stackexchange.com/q/470983/138900 — AndréC, Jul 12 '19 at 19:04
@AndréC This version's transparent cubes will work very nicely with the "folding" part in the bottom two subfigures. Thank you :) — anonymous_pigeon, Jul 12 '19 at 19:12
Remember that you have the right to answer your own question. If you find a solution, don't forget to publish it. https://tex.stackexchange.com/help/self-answer :) — AndréC, Jul 12 '19 at 19:17
@siracusa I'm very sorry for posting this question without having tried to solve it first. I want to make an attempt and update the post somehow. In the meantime, should I delete this question and post an updated version with a first attempt? Can I just edit this question and add the code? I don't know what the best policy is here :) — anonymous_pigeon, Jul 12 '19 at 19:17
Do not delete this question, the site you have linked to is exciting. Answer your own question, it will be a good exercise. — AndréC, Jul 12 '19 at 19:21
There's no need to delete it. It's perfectly fine to edit your question and add your current attempt version. Giving people a base they can work on will increase your chances of getting a good answer. — siracusa, Jul 12 '19 at 19:41
@AndréC Well, when you say "This is possible with the construction of Jang Soo Kim", can you spell this out? BTW, Jang Soo Kim's code works, it does not have to get "fixed". — , Jul 13 '19 at 04:56
@AndréC It is not possible with (jus) these routines, you would have to modify them drastically, simply because there is no real 3D ordering in it. I am talking specifically about the upper left part of the figure with the three stacks of cubes. — , Jul 13 '19 at 05:28

score 12 · Answer 1 · 2019-07-13T07:26:43.260

This is a version that uses orthographic projections. That is, you can adjust the view angles, at least to some extent, and the view is realistic (in the limit of a distant observer). Everything is stored in pgf keys, so highly adjustable. The styles and keys are essentially taken from this answer and this answer, where you can find animations that illustrate what it means that the view is adjustable.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\tikzset{plane/.style n args={3}{insert path={%
#1 -- ++ #2 -- ++ #3 -- ++ ($-1*#2$) -- cycle}},
unit xy plane/.style={plane={#1}{(\CubeX,0,0)}{(0,\CubeY,0)}},
unit xz plane/.style={plane={#1}{(\CubeX,0,0)}{(0,0,\CubeZ)}},
unit yz plane/.style={plane={#1}{(0,\CubeY,0)}{(0,0,\CubeZ)}},
get projections/.style={insert path={%
let \p1=(1,0,0),\p2=(0,1,0)  in 
[/utils/exec={\pgfmathtruncatemacro{\xproj}{sign(\x1)}\xdef\xproj{\xproj}
\pgfmathtruncatemacro{\yproj}{sign(\x2)}\xdef\yproj{\yproj}
\pgfmathtruncatemacro{\zproj}{sign(cos(\tdplotmaintheta))}\xdef\zproj{\zproj}}]}},
pics/unit cube/.style={code={
\path[get projections];
\ifnum\zproj=-1
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\yproj=1
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\xproj=1
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\zproj>-1
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,\CubeZ/2)}]; 
\fi
}},
3d cube/.cd,
xy face/.style={fill=red!20},
xz face/.style={fill=blue!20},
yz face/.style={fill=orange!30},
num cubes x/.estore in=\NumCubesX,
num cubes y/.estore in=\NumCubesY,
num cubes z/.estore in=\NumCubesZ,
num cubes x/.initial=1,num cubes y/.initial=1,num cubes z/.initial=1,
cube x/.estore in=\CubeX,
cube y/.estore in=\CubeY,
cube z/.estore in=\CubeZ,
cube x=1,cube y=1,cube z=1,
cube scale/.initial=1,
every face/.style={draw,very thick},
/tikz/pics/.cd,
cube array/.style={code={%
 \tikzset{3d cube/.cd,#1}
 %\typeout{\NumCubesX,\NumCubesY,\NumCubesZ}
  \path[get projections];
  \ifnum\yproj=1
   \def\LstX{1,...,\NumCubesX}
  \else 
   \ifnum\NumCubesX>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesX-1}
    \def\LstX{\NumCubesX,\NextToLast,...,1}
   \else
    \def\LstX{1}   
   \fi 
  \fi
  \ifnum\xproj=-1
   \def\LstY{1,...,\NumCubesY}
  \else 
   \ifnum\NumCubesY>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesY-1}
    \def\LstY{\NumCubesY,\NextToLast,...,1}
   \else
    \def\LstY{1}   
   \fi 
  \fi
  \ifnum\zproj=1
   \def\LstZ{1,...,\NumCubesZ}
  \else 
   \ifnum\NumCubesZ>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesZ-1}
    \def\LstZ{\NumCubesZ,\NextToLast,...,1}
   \else
    \def\LstZ{1}   
   \fi 
  \fi
  \foreach \X in \LstX
  {\foreach \Y in \LstY
   {\foreach \Z in \LstZ
    {\path (\X-\NumCubesX/2-1/2,\Y-\NumCubesY/2-1/2,\Z-\NumCubesY/2-1/2)
      pic[scale=\pgfkeysvalueof{/tikz/3d cube/cube scale}]{unit cube};}}
  } 
}}
}

\begin{document}
\tdplotsetmaincoords{70}{50} % the first argument cannot be larger than 90
\begin{tikzpicture}[line join=round,tdplot_main_coords,font=\sffamily,3d cube/.cd,
 num cubes x=1,num cubes y=1,num cubes z=1]
\begin{scope} 
% top left
 \path (-3,0,0) pic{cube array={num cubes y=3,num cubes z=3}}
 ++(1,-1/2,1/2) pic{cube array={num cubes y=2,num cubes z=2}}
 ++(1,-1/2,1/2) pic{cube array={num cubes y=1,num cubes z=1}};
% top right
 \path (3,3,0) pic{cube array={num cubes x=3,num cubes z=3}}
  ++ (1/2,-1,1) pic{cube array={num cubes x=2,num cubes z=2}}
  ++ (1/2,-1,1) pic{cube array={num cubes y=1,num cubes z=1}};
% bottom
 \path (0,2,-3) pic{cube array={num cubes x=3,num cubes y=3}}
   ++ (1/2,-1/2,1/2) pic{cube array={num cubes x=2,num cubes y=2}}
   ++ (1/2,-1/2,1/2) pic{cube array={num cubes x=1,num cubes y=1}};
\end{scope}
\begin{scope}[xshift=11cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}}
 (1/2,1,1) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \draw[line width=2pt] (-1,-1.5,-5/2) -- ++ (0,0,1)
 -- ++ (1,0,0) -- ++ (0,0,1)  -- ++ (1,0,0) -- ++ (0,0,1)
 -- ++ (1,0,0) -- ++ (0,1,0) -- ++ (0,0,-1) -- ++ (0,1,0) -- ++ (0,0,-1)
 -- ++ (0,1,0) -- ++ (0,0,-1);
\end{scope}
\tikzset{darker/.style={3d cube/.cd,cube z=1/2,xy face/.style={fill=red!40},
    xz face/.style={fill=blue!40},yz face/.style={fill=orange!60}}}
\begin{scope}[yshift=-8cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}};
 \path[3d cube/cube z=1/2]
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \path[3d cube/.cd,cube z=1/2]
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \path[darker]
 (1/2,1,5/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
\end{scope}
\begin{scope}[yshift=-8cm,xshift=11cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}};
 \path[3d cube/.cd,cube z=1/2]
 (-3/2,1,3/4) pic[darker]{cube array={num cubes x=1,num cubes y=1,num cubes z=1}}
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 (-1,0,3/4) pic[darker]{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 (1,0,3/4) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 (-1/2,-1,3/4) pic[darker]{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 (3/2,-1,3/4) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
\end{scope}
\end{tikzpicture}
\end{document}

In order to allow for a nontrivial opacity, one needs also to draw the hidden faces because they will unhide if the faces covering them become transparent.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\tikzset{plane/.style n args={3}{insert path={%
#1 -- ++ #2 -- ++ #3 -- ++ ($-1*#2$) -- cycle}},
unit xy plane/.style={plane={#1}{(\CubeX,0,0)}{(0,\CubeY,0)}},
unit xz plane/.style={plane={#1}{(\CubeX,0,0)}{(0,0,\CubeZ)}},
unit yz plane/.style={plane={#1}{(0,\CubeY,0)}{(0,0,\CubeZ)}},
get projections/.style={insert path={%
let \p1=(1,0,0),\p2=(0,1,0)  in 
[/utils/exec={\pgfmathtruncatemacro{\xproj}{sign(\x1)}\xdef\xproj{\xproj}
\pgfmathtruncatemacro{\yproj}{sign(\x2)}\xdef\yproj{\yproj}
\pgfmathtruncatemacro{\zproj}{sign(cos(\tdplotmaintheta))}\xdef\zproj{\zproj}}]}},
pics/unit cube/.style={code={
\path[get projections];
\ifnum\zproj=-1
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else 
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,\CubeZ/2)}]; 
\fi
\ifnum\yproj=1
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\xproj=1
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\yproj=1
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/yz face,unit yz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\xproj=1
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\else
 \path[3d cube/every face,3d cube/xz face,unit xz plane={(-\CubeX/2,\CubeY/2,-\CubeZ/2)}]; 
\fi
\ifnum\zproj>-1
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,\CubeZ/2)}]; 
\else 
 \path[3d cube/every face,3d cube/xy face,unit xy plane={(-\CubeX/2,-\CubeY/2,-\CubeZ/2)}]; 
\fi
}},
3d cube/.cd,
xy face/.style={fill=red!20},
xz face/.style={fill=blue!20},
yz face/.style={fill=orange!30},
num cubes x/.estore in=\NumCubesX,
num cubes y/.estore in=\NumCubesY,
num cubes z/.estore in=\NumCubesZ,
num cubes x/.initial=1,num cubes y/.initial=1,num cubes z/.initial=1,
cube x/.estore in=\CubeX,
cube y/.estore in=\CubeY,
cube z/.estore in=\CubeZ,
cube x=1,cube y=1,cube z=1,
cube scale/.initial=1,
every face/.style={draw,very thick},
/tikz/pics/.cd,
cube array/.style={code={%
 \tikzset{3d cube/.cd,#1}
 %\typeout{\NumCubesX,\NumCubesY,\NumCubesZ}
  \path[get projections];
  \ifnum\yproj=1
   \def\LstX{1,...,\NumCubesX}
  \else 
   \ifnum\NumCubesX>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesX-1}
    \def\LstX{\NumCubesX,\NextToLast,...,1}
   \else
    \def\LstX{1}   
   \fi 
  \fi
  \ifnum\xproj=-1
   \def\LstY{1,...,\NumCubesY}
  \else 
   \ifnum\NumCubesY>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesY-1}
    \def\LstY{\NumCubesY,\NextToLast,...,1}
   \else
    \def\LstY{1}   
   \fi 
  \fi
  \ifnum\zproj=1
   \def\LstZ{1,...,\NumCubesZ}
  \else 
   \ifnum\NumCubesZ>1
    \pgfmathtruncatemacro{\NextToLast}{\NumCubesZ-1}
    \def\LstZ{\NumCubesZ,\NextToLast,...,1}
   \else
    \def\LstZ{1}   
   \fi 
  \fi
  \foreach \X in \LstX
  {\foreach \Y in \LstY
   {\foreach \Z in \LstZ
    {\path (\X-\NumCubesX/2-1/2,\Y-\NumCubesY/2-1/2,\Z-\NumCubesY/2-1/2)
      pic[scale=\pgfkeysvalueof{/tikz/3d cube/cube scale}]{unit cube};}}
  } 
}}
}

\begin{document}
\tdplotsetmaincoords{70}{50} % the first argument cannot be larger than 90
\begin{tikzpicture}[line join=round,tdplot_main_coords,font=\sffamily,3d cube/.cd,
 num cubes x=1,num cubes y=1,num cubes z=1]
\begin{scope} 
% top left
 \path (-3,0,0) pic{cube array={num cubes y=3,num cubes z=3}}
 ++(1,-1/2,1/2) pic{cube array={num cubes y=2,num cubes z=2}}
 ++(1,-1/2,1/2) pic{cube array={num cubes y=1,num cubes z=1}};
% top right
 \path (3,3,0) pic{cube array={num cubes x=3,num cubes z=3}}
  ++ (1/2,-1,1) pic{cube array={num cubes x=2,num cubes z=2}}
  ++ (1/2,-1,1) pic{cube array={num cubes y=1,num cubes z=1}};
% bottom
 \path (0,2,-3) pic{cube array={num cubes x=3,num cubes y=3}}
   ++ (1/2,-1/2,1/2) pic{cube array={num cubes x=2,num cubes y=2}}
   ++ (1/2,-1/2,1/2) pic{cube array={num cubes x=1,num cubes y=1}};
\end{scope}
\begin{scope}[xshift=11cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}}
 (1/2,1,1) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \draw[line width=2pt] (-1,-1.5,-5/2) -- ++ (0,0,1)
 -- ++ (1,0,0) -- ++ (0,0,1)  -- ++ (1,0,0) -- ++ (0,0,1)
 -- ++ (1,0,0) -- ++ (0,1,0) -- ++ (0,0,-1) -- ++ (0,1,0) -- ++ (0,0,-1)
 -- ++ (0,1,0) -- ++ (0,0,-1);
\end{scope}
\tikzset{darker/.style={3d cube/.cd,cube z=1/2,xy face/.style={fill=red!40},
    xz face/.style={fill=blue!40},yz face/.style={fill=orange!60}}}
\begin{scope}[yshift=-8cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}};
 \path[3d cube/cube z=1/2]
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \path[3d cube/.cd,cube z=1/2]
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
 \path[darker,/tikz/opacity=0.6]
 (1/2,1,5/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 ++(1/2,-1,0) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
\end{scope}
\begin{scope}[yshift=-8cm,xshift=11cm] 
% top left
 \path (0,0,-1) pic{cube array={num cubes x=4,num cubes y=3,num cubes z=3}};
 \path[3d cube/.cd,cube z=1/2,/tikz/opacity=0.6]
 (-3/2,1,3/4) pic[darker]{cube array={num cubes x=1,num cubes y=1,num cubes z=1}}
 (1/2,1,3/4) pic{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 (-1,0,3/4) pic[darker]{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 (1,0,3/4) pic{cube array={num cubes x=2,num cubes y=1,num cubes z=1}}
 (-1/2,-1,3/4) pic[darker]{cube array={num cubes x=3,num cubes y=1,num cubes z=1}}
 (3/2,-1,3/4) pic{cube array={num cubes x=1,num cubes y=1,num cubes z=1}};
\end{scope}
\end{tikzpicture}
\end{document}

score 10 · Answer 2 · edited Jul 14 '19 at 14:13

Here is an answer to show: we can very well use Jang Soo Kim's code routine to replicate this proof without words.

Here is the result, the replication is not accurate because the angles of this perspective spoil the perspective effect.

So without changing the routine, I modified the angles and length of some faces to get a better 3D rendering.

Here is the result:

Here is the code of the figure generated with the Jang Soo Kim code modified by myself in this answer: https://tex.stackexchange.com/a/470985/138900 (image at the top of these page)

\documentclass[tikz,border=5mm]{standalone}
% 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!30,fill opacity=1, 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=cyan!50,fill opacity=1, 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=yellow!40!green!80!white,fill opacity=1, 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[2][0]{
 \setcounter{x}{-1}
  \foreach \a in {#2} {
        \addtocounter{x}{1}
        \setcounter{y}{-1}
            \foreach \b in \a {
            \addtocounter{y}{1}
            \setcounter{z}{-1}
            \addtocounter{z}{#1} %partition of the desired floor (layer)
            \ifnum \b>0
            \foreach \c in {1,...,\b} {
                \addtocounter{z}{1}
                \cube{\value{x}}{\value{y}}{\value{z}}
      }\fi
    }
  }
}
\begin{document} 


\begin{tikzpicture}
\planepartition{{3},{3},{3}}% Old syntax is functional
\planepartition[1]{{0,0},{0,2},{0,2}}
\planepartition[2]{{0},{0},{0,0,1}}

\begin{scope}[xshift=5cm]
\planepartition{{3,3,3}}% Old syntax is functional
\planepartition[1]{{0},{0,2,2}}
\planepartition[2]{{0},{0},{0,0,1}}
\end{scope}

\begin{scope}[shift={(3,-4)}]
\planepartition{{1},{1},{1}}
\planepartition{{0,1},{0,2},{0,2}}
\planepartition{{0,0,1},{0,0,2},{0,0,3}}
\end{scope}
\end{tikzpicture}
\end{document}

The algorithm (called routine by marmot), i. e. the code of the command \planepartition has not been modified.

\newcommand\planepartition[2][0]{
 \setcounter{x}{-1}
  \foreach \a in {#2} {
        \addtocounter{x}{1}
        \setcounter{y}{-1}
            \foreach \b in \a {
            \addtocounter{y}{1}
            \setcounter{z}{-1}
            \addtocounter{z}{#1} %partition of the desired floor (layer)
            \ifnum \b>0
            \foreach \c in {1,...,\b} {
                \addtocounter{z}{1}
                \cube{\value{x}}{\value{y}}{\value{z}}
      }\fi
    }
  }
}

Only the initial conditions (angles and face length) have been modified. The calls to the routine (\planepartition) are identical in both versions:

\begin{tikzpicture}
\planepartition{{3},{3},{3}}% Old syntax is functional
\planepartition[1]{{0,0},{0,2},{0,2}}
\planepartition[2]{{0},{0},{0,0,1}}

\begin{scope}[xshift=5cm]
\planepartition{{3,3,3}}% Old syntax is functional
\planepartition[1]{{0},{0,2,2}}
\planepartition[2]{{0},{0},{0,0,1}}
\end{scope}

\begin{scope}[shift={(3,-4)}]
\planepartition{{1},{1},{1}}
\planepartition{{0,1},{0,2},{0,2}}
\planepartition{{0,0,1},{0,0,2},{0,0,3}}
\end{scope}
\end{tikzpicture}

Complete code:

\documentclass[tikz,border=5mm]{standalone}

\newcounter{x}
\newlength{\x}
\setlength{\x}{.8cm}
\newcounter{y}
\newcounter{z}
% The angles of x,y,z-axes
\def\xangle{30}
\def\yangle{10}
\newcommand\xaxis{180+\xangle}
\newcommand\yaxis{-\yangle}
\newcommand\zaxis{90}
% The top side of a cube
\newcommand\topside[3]{
  \fill[fill=yellow!60,fill opacity=1, draw=black,shift={(\xaxis:#1\x)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (\xangle:\x) --++ (180-\yangle:1) --++(180+\xangle:\x)--cycle;
}
% The left side of a cube
\newcommand\leftside[3]{
  \fill[fill=orange!60,fill opacity=1, draw=black,shift={(\xaxis:#1\x)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (0,-1) --++ (180-\yangle:1) --(180-\yangle:1)--(0,0);
}
% The right side of a cube
\newcommand\rightside[3]{
  \fill[fill=blue!60,fill opacity=1, draw=black,shift={(\xaxis:#1\x)},shift={(\yaxis:#2)},
  shift={(\zaxis:#3)}] (0,0) -- (\xangle:\x) --++ (0,-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[2][0]{
 \setcounter{x}{-1}
  \foreach \a in {#2} {
        \addtocounter{x}{1}
        \setcounter{y}{-1}
            \foreach \b in \a {
            \addtocounter{y}{1}
            \setcounter{z}{-1}
            \addtocounter{z}{#1} %partition of the desired floor (layer)
            \ifnum \b>0
            \foreach \c in {1,...,\b} {
                \addtocounter{z}{1}
                \cube{\value{x}}{\value{y}}{\value{z}}
      }\fi
    }
  }
}
\begin{document} 



\begin{tikzpicture}
\planepartition{{3},{3},{3}}% Old syntax is functional
\planepartition[1]{{0,0},{0,2},{0,2}}
\planepartition[2]{{0},{0},{0,0,1}}

\begin{scope}[xshift=5cm]
\planepartition{{3,3,3}}% Old syntax is functional
\planepartition[1]{{0},{0,2,2}}
\planepartition[2]{{0},{0},{0,0,1}}
\end{scope}

\begin{scope}[shift={(3,-4)}]
\planepartition{{1},{1},{1}}
\planepartition{{0,1},{0,2},{0,2}}
\planepartition{{0,0,1},{0,0,2},{0,0,3}}
\end{scope}
\end{tikzpicture}
\end{document}

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Replicating this Visual Proof with TikZ

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