我怎样才能更快地绘制条形和星形？

Question

我想使用条形和星号方法列出方程的非负解的数量x + y + z = 4。我尝试过

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fourier}
\begin{document}
\begin{tabular}{|c|c|}
\hline 
$(0, 0, 4)$ & $\big|\big|\star \star \star\star $ \\ 
\hline 
$(0, 1, 3)$ & $\big|\star\big| \star \star\star $ \\ 
\hline 
$(0, 2, 2)$ & $\big|\star\star \big|  \star\star $ \\ 
\hline 
$(0, 3, 1)$ & $\big|\star\star \star \big|  \star $ \\ 
\hline 
$(0, 4, 0)$ & $\big|\star\star \star  \star\big|  $ \\ 
\hline 
$(1, 0, 3)$ & $\star\big|\big|\star \star  \star  $ \\ 
\hline 
$(1, 1, 2)$ & $\star\big|\star \big|\star   \star  $ \\ 
\hline 
$(1, 2, 1)$ & $\star\big|\star \star\big|   \star  $ \\ 
\hline 
$(1, 3, 0)$ & $\star\big|\star \star   \star \big| $ \\ 
\hline 
$(2, 0, 2)$ & $\star \star\big|\big|\star   \star  $ \\ 
\hline 
$(2, 1, 1)$ & $\star \star\big|\star\big|   \star  $ \\ 
\hline 
$(2, 2, 0)$ & $\star \star\big|\star \star\big|     $ \\ 
\hline 
$(3, 0, 1)$ & $\star \star\star \big|\big| \star    $ \\ 
\hline 
$(3, 1, 0)$ & $\star \star\star \big| \star \big|   $ \\ 
\hline 
$(4, 0, 0)$ & $\star \star\star \star\big|  \big|   $ \\ 
\hline 
\end{tabular} 
\end{document}

我知道，该解的非负解的数量x + y + z = 4是C_(4+3-1)^4=15。有没有什么方法可以更快地表示条形和星形？

Answer 1

强制expl3解决方案：两个嵌套循环来构建表体。

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\starsandbars}{m}
 {
  \minthao_starsandbars:n { #1 }
 }

\tl_new:N \l__minthao_table_body_tl

\cs_new_protected:Npn \minthao_starsandbars:n #1
 {
  \tl_clear:N \l__minthao_table_body_tl
  \int_step_inline:nnnn { 0 } { 1 } { #1 } % ##1 is x
   {
    \int_step_inline:nnnn { 0 } { 1 } { #1 - ##1 } % ####1 is y
     {
      \tl_put_right:Nx \l__minthao_table_body_tl
       {
        $(\int_to_arabic:n { ##1 },
          \int_to_arabic:n { ####1 },
          \int_to_arabic:n { #1 - ##1 - ####1 })$ &
       }
      \tl_put_right:Nx \l__minthao_table_body_tl
       {
        \prg_replicate:nn { ##1 } { $\star$ }
        \exp_not:n { \,\vline\, }
        \prg_replicate:nn { ####1 } { $\star$ }
        \exp_not:n { \,\vline\, }
        \prg_replicate:nn { #1 - ##1 - ####1 } { $\star$ }
       }
      \tl_put_right:Nn \l__minthao_table_body_tl { \\ \hline }
     }
   }
  \begin{tabular}{|c|c|}
    \hline
    \l__minthao_table_body_tl
  \end{tabular}
 }
\ExplSyntaxOff

\begin{document}

\starsandbars{4}

\end{document}

具有键值接口的扩展版本；键是

• longtable（布尔值）；如果表达，则longtable产生a
• preamble，用于指定longtable前导码\endhead等，或者在表的开头设置一些内容（参见第二个示例）

图片仅显示前两页。

\documentclass{article}
\usepackage{xparse,longtable}

\ExplSyntaxOn
\NewDocumentCommand{\starsandbars}{O{}m}
 {
  \group_begin:
  \keys_set:nn { minthao/starsandbars } { #1 }
  \minthao_starsandbars:n { #2 }
  \__minthao_starsandbars_print_table:
  \group_end:
 }

\keys_define:nn { minthao/starsandbars }
 {
  longtable .bool_set:N = \l__minthao_starsandbars_longtable_bool,
  longtable .default:n = { true },
  longtable .initial:n = { false },
  preamble  .tl_set:N  = \l__minthao_starsandbars_preamble_tl,
  preamble  .initial:n = \hline,
 }

\tl_new:N \l__minthao_table_body_tl

\cs_new_protected:Npn \minthao_starsandbars:n #1
 {
  \tl_clear:N \l__minthao_table_body_tl
  \int_step_inline:nnnn { 0 } { 1 } { #1 } % ##1 is x
   {
    \int_step_inline:nnnn { 0 } { 1 } { #1 - ##1 } % ####1 is y
     {
      \tl_put_right:Nx \l__minthao_table_body_tl
       {
        $(\int_to_arabic:n { ##1 },
          \int_to_arabic:n { ####1 },
          \int_to_arabic:n { #1 - ##1 - ####1 })$ &
       }
      \tl_put_right:Nx \l__minthao_table_body_tl
       {
        \prg_replicate:nn { ##1 } { $\star$ }
        \exp_not:n { \,\vline\, }
        \prg_replicate:nn { ####1 } { $\star$ }
        \exp_not:n { \,\vline\, }
        \prg_replicate:nn { #1 - ##1 - ####1 } { $\star$ }
       }
      \tl_put_right:Nn \l__minthao_table_body_tl { \\ \hline }
     }
   }
 }

\cs_new_protected:Npn \__minthao_starsandbars_print_table:
 {
  \bool_if:NTF \l__minthao_starsandbars_longtable_bool
   { \begin{longtable} }
   { \begin{tabular} }
   {|c|c|}
  \l__minthao_starsandbars_preamble_tl
  \l__minthao_table_body_tl
  \bool_if:NTF \l__minthao_starsandbars_longtable_bool
   { \end{longtable} }
   { \end{tabular} }
 }



\ExplSyntaxOff

\begin{document}

\starsandbars{3}

\bigskip

\starsandbars[
  preamble=\hline\multicolumn{2}{|c|}{$x+y+z=4$}\\\hline
]{4}

\starsandbars[
  longtable,
  preamble={
    \caption{$x+y+z=12$}\\
    \hline
    \endfirsthead
    \caption*{$x+y+z=12$ (continued)}\\
    \hline
    \endhead
  }
]{12}

\end{document}

Answer 2

不确定是否要求在括号中显示数字和画星号和横线。这个\starsandbars宏只画星号和横线：

\documentclass[varwidth,border=5]{standalone}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fourier}
\usepackage{pgffor}

\def\starsandbars#1{{\ensuremath{%
  \edef\starlist{#1}%
  \foreach \i [count=\a] in \starlist{%
    \ifnum\a>1\big|\fi%
    \ifnum\i>0\foreach \j in {1,...,\i}{\star}\fi%
  }}}}%
\begin{document}
\foreach \i in {{0,1,2},{1,2,1},{0,0,5}, {1,3,0}, {4,3,2,6}}{\leavevmode
\hbox to .75in{(\i)\hfil} \starsandbars{\i} \par
}
\end{document}

Answer 3

问题已被标记，但如果您想按照原始表述tikz-pgf来做，那么这是一项理想的工作。tabular\xintFor

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fourier}

\usepackage{xinttools}
% we need _ as private letter because we will employ \xint_dothis
% and \xint_orthat

\catcode`_ 11
\newcommand\barsandstars [1]{% solving x+y+z=#1 in non-negative integers
  \edef\tmp_N {\numexpr\the\numexpr #1\relax\relax}%
  \def\tmp_a {0}%
  \def\tmp_b {0}%
  \edef\tmp_c {\the\tmp_N }%
  \begin{tabular}{|c|c|}
   \hline
   \xintFor ##1 in {\xintegers [0+1]}\do 
   {%  ##1 = insertion location of first bar
       \xintFor ##2 in {\xintegers [{\numexpr##1+1\relax}+1]}\do 
       {%  ##2 = insertion location of second bar
        %     (first bar insertion has shifted by one all things to its right)
           $(\tmp_a,\tmp_b,\tmp_c)$
           &
           $\xintFor ##3 in {\xintegers [0+1]}\do 
            {%  
               \ifnum ##3<##1\xint_dothis{\star}\fi
               \ifnum ##3=##1\xint_dothis{\big|}\fi
               \ifnum ##3<##2\xint_dothis{\star}\fi
               \ifnum ##3=##2\xint_dothis{\big|}\fi
               \xint_orthat {\star}%
               \ifnum ##3>\tmp_N \expandafter\xintBreakFor\fi
            }$
% The trick is to always decide if we need to continue BEFORE inserting \\
% and we must do non-expandable things before the \\
           \xdef\tmp_b {\the\numexpr\tmp_b+1}%
           \xdef\tmp_c {\the\numexpr\tmp_c-1}%
           \ifnum ##2>\tmp_N \expandafter\xintBreakFor\fi
           \\\hline
       }%
       \xdef\tmp_a {\the\numexpr\tmp_a+1\relax }%
       \gdef\tmp_b {0}%
       \xdef\tmp_c {\the\numexpr \tmp_N-\tmp_a\relax}%
       \ifnum ##1=\tmp_N \expandafter\xintBreakFor\fi
       \\\hline
   }%
   \\\hline
  \end{tabular} 
}  
\catcode`_8


\begin{document}
  \barsandstars {7}
\end{document}

Answer 4

如果“更快”意味着更短的代码不仅适用于 4，那么这里有一个 TikZ 解决方案：

\documentclass[tikz,border=7mm]{standalone}
\begin{document}
\begin{tikzpicture}[scale=.7]
  \def\total{4} % i+j+k = total
  \xdef\y{0}
  \foreach \i in {0,...,\total}{
    \pgfmathsetmacro{\maxj}{int(\total-\i)}
    \foreach \j in {0,...,\maxj}{
      \pgfmathparse{int(\y+1)}\xdef\y{\pgfmathresult}
      \pgfmathsetmacro{\k}{int(\total-\i-\j)}
      \foreach \i in {1,...,\total}
        \node at (\i,\y){$\star$};
      \node at (.45+\j+\i,\y) {$\big|$};
      \node at (.55+\i,\y) {$\big|$};
      \node[left] at (-1,\y){(\i,\j,\k)};
    }
  }
\end{tikzpicture}
\end{document}

如果你设置，\def\total{2}你会得到：

我不会公布 7 的结果；)