简化 PSTricks 代码

Question

考虑以下代码：

\documentclass{article}

\usepackage[
  a4paper,
  landscape,
  margin = 4cm
]{geometry}
\usepackage{xfp}
\usepackage{pstricks}
\psset{dimen = m}
\usepackage{multido}

\begin{document}

\begin{pspicture}(21,8)
  \multido
    {\i = 0+6}
    {4}
    {\psframe(\i,0)(\fpeval{\i+3},2)}
  \multido
    {\i = 3+6}
    {3}
    {\psframe(\i,2)(\fpeval{\i+3},4)
     \psframe[fillstyle = solid, fillcolor = gray](\i,0)(\fpeval{\i+3},2)
     \rput(\fpeval{\i+1.5},1){\Large $+$}
     \psline{->}(\fpeval{\i-0.2},1.8)(\fpeval{\i+0.2},2.2)
     \psline{->}(\fpeval{\i+3.2},1.8)(\fpeval{\i+2.8},2.2)}
  \multido
    {\i = 6+6}
    {2}
    {\psframe(\i,4)(\fpeval{\i+3},6)
     \psframe[fillstyle = solid, fillcolor = gray](\i,2)(\fpeval{\i+3},4)
     \rput(\fpeval{\i+1.5},3){\Large $+$}
     \psline{->}(\fpeval{\i-0.2},3.8)(\fpeval{\i+0.2},4.2)
     \psline{->}(\fpeval{\i+3.2},3.8)(\fpeval{\i+2.8},4.2)}
  \multido
    {\i = 9+6}
    {1}
    {\psframe(\i,6)(\fpeval{\i+3},8)
     \psframe[fillstyle = solid, fillcolor = gray](\i,4)(\fpeval{\i+3},6)
     \rput(\fpeval{\i+1.5},5){\Large $+$}
     \psline{->}(\fpeval{\i-0.2},5.8)(\fpeval{\i+0.2},6.2)
     \psline{->}(\fpeval{\i+3.2},5.8)(\fpeval{\i+2.8},6.2)}
\end{pspicture}

\end{document}

\multido我确信通过在 a 中使用 a 可以更“优雅”地完成此操作，\multido但我做不到。我该怎么做？

更新

这是我最终得到的结果，感谢 user187802 的出色回答：

\documentclass[a4paper]{article}

\usepackage[margin = 1cm]{geometry}
\usepackage{pstricks}
\psset{dimen = m}
\usepackage{multido}
\usepackage{xfp}

% Parameters
\def\pilStr{0.2}      % arrow size
\def\klodsBredde{2.5} % block width
\def\klodsHoejde{1.6} % block height
\def\pyramideStr{4}   % pyramide size (natural number)

\newcommand*\pil{
  \psline{->}(-\pilStr,-\pilStr)(\pilStr,\pilStr)
}
\def\kasseTom(#1,#2){
  \psframe(#1,#2)
}
\def\kassePlus(#1,#2){
  \psframe[
    fillstyle = solid,
    fillcolor = gray!75
  ](#1,#2)
  \rput(\fpeval{#1/2},\fpeval{#2/2}){\large$+$}
  \rput(0,#2){\pil}
  \rput{90}(#1,#2){\pil}
}
\newcommand*\pyramideBredde{\fpeval{(2*\pyramideStr-1)*\klodsBredde}}
\newcommand*\pyramideHoejde{\fpeval{\klodsHoejde*\pyramideStr}}
\makeatletter
\newcommand*\pyramide{
  \@ifnextchar(\psPyramide{\psPyramide(2,1)}
}
\def\psPyramide(#1,#2)#3{
    \multido
      {\rX = 0+#1,
       \rY = 0+#2,
       \iZ = 0+1}
      {#3}
      {\multido
         {\rA = \rX+\fpeval{2*#1}}
         {\fpeval{#3-\iZ}}
         {\rput(\rA,\rY){\kasseTom(#1,#2)}}
       \multido
         {\rB = \fpeval{\rX+#1}+\fpeval{2*#1}}
         {\fpeval{#3-\iZ-1}}
         {\rput(\rB,\rY){\kassePlus(#1,#2)}}}
}
\makeatother
\def\udtryk(#1,#2)#3{
  \rput(\fpeval{(2*#1+#2-2.5)*\klodsBredde},\fpeval{(#2-0.5)*\klodsHoejde}){#3}
}

\begin{document}

\begin{figure}
  \centering
  \begin{pspicture}(\pyramideBredde,\pyramideHoejde)
    \pyramide(\klodsBredde,\klodsHoejde){\pyramideStr}
    \udtryk(1,1){$x$}
    \udtryk(2,1){$x$}
    \udtryk(3,1){$y$}
    \udtryk(4,1){$y$}
    \udtryk(1,2){\textcolor{red}{$2x$}}
    \udtryk(2,2){$x + y$}
    \udtryk(3,2){\textcolor{red}{$2y$}}
    \udtryk(1,3){\textcolor{red}{$3x + y$}}
    \udtryk(2,3){\textcolor{red}{$x + 3y$}}
    \udtryk(1,4){\textcolor{red}{$4x + 4y$}}
  \end{pspicture}
  \caption{An addition pyramide.}
  \label{fig:add-pyramide}
\end{figure}

\end{document}

Answer 1

运行lualatex：

\documentclass{article}
\usepackage[a4paper,margin = 1cm]{geometry}
\usepackage{pstricks}
\psset{dimen = m}
\usepackage{multido}

\def\Arrow{\psline{->}(-0.2,-0.2)(0.2,0.2)}
\def\Frame#1(#2,#3){%
  \psframe#1[linecolor=black!50](#2,#3)%
  \if$#1$\else\rput(!#2 2 div #3 2 div){+}\rput(0,#3){\Arrow}\rput{90}(#2,#3){\Arrow}\fi}
\makeatletter
\def\Bricks{\@ifnextchar(\psBricks{\psBricks(2,1)}}%
\def\psBricks(#1,#2)#3{%
  \begin{pspicture}[calcframe]
  \edef\dX{\numexpr2*#1}%
  \edef\dY{\numexpr2*#2}%
  \multido{\iX=0+#1,\iY=0+#2,\iZ=0+1}{#3}{%
    \multido{\iA=\iX+\dX}{\numexpr#3-\iZ}{\rput(\iA,\iY){\Frame(#1,#2)}}%
    \multido{\iB={\numexpr\iX+#1}+\dX}{\numexpr#3-\iZ-1}{\rput(\iB,\iY){\Frame*(#1,#2)}}%
  }
\end{pspicture}}
\makeatother
\begin{document}

\psset{unit=0.5}

\Bricks(2,1){4}

\Bricks(3,2){4}

\Bricks{6}

\Bricks(1,1){4}

\end{document}