自动生成乘法表并指定颜色的行、列和对角线

Question 1

本页上各种答案和 OP 代码的混合片段......

\documentclass[pstricks,border=5]{standalone}    
\usepackage{expl3}
\ExplSyntaxOn
  \cs_new_eq:NN \calc \fp_eval:n
\ExplSyntaxOff    
\psset{dimen = m, fillstyle = solid}
\begin{document} 
\def\N{11}
\begin{pspicture}(\calc{\N+1},\calc{\N+1})
\psframe[fillcolor = orange](!0 \N\space 1 add)(1,\N)
\rput(!0.5 \N\space 0.5 add){$\cdot$}     
\foreach \i in {1,...,\N}{
  \psframe[fillcolor = orange](\i, \N)(\calc{\i+1}, \calc{\N+1})
  \rput(\calc{\i+.5}, \calc{\N+.5}){$\i$}
  \psframe[fillcolor = orange](0, \calc{\N-\i})(1, \calc{\N-\i+1})
  \rput(0.5, \calc{\N-\i+.5}){$\i$}
}     
\foreach \iA in {1,...,\N}{\foreach \iB in {1,...,\N}{
    \edef\x{\calc{abs(\iA-\iB)}} 
    \ifnum\x=5
      \colorlet{fill color}{blue!20}
    \else
      \colorlet{fill color}{white}
    \fi
    \psframe[fillcolor=fill color](\iA, \calc{\N-\iB})(\calc{\iA+1}, \calc{\N-\iB+1})
    \rput(\calc{\iA+.5}, \calc{\N-\iB+.5}){$\calc{\iA*\iB}$} 
}}
\end{pspicture}    
\end{document}

在此处输入图片描述

Answer

本页上各种答案和 OP 代码的混合片段......

\documentclass[pstricks,border=5]{standalone}    
\usepackage{expl3}
\ExplSyntaxOn
  \cs_new_eq:NN \calc \fp_eval:n
\ExplSyntaxOff    
\psset{dimen = m, fillstyle = solid}
\begin{document} 
\def\N{11}
\begin{pspicture}(\calc{\N+1},\calc{\N+1})
\psframe[fillcolor = orange](!0 \N\space 1 add)(1,\N)
\rput(!0.5 \N\space 0.5 add){$\cdot$}     
\foreach \i in {1,...,\N}{
  \psframe[fillcolor = orange](\i, \N)(\calc{\i+1}, \calc{\N+1})
  \rput(\calc{\i+.5}, \calc{\N+.5}){$\i$}
  \psframe[fillcolor = orange](0, \calc{\N-\i})(1, \calc{\N-\i+1})
  \rput(0.5, \calc{\N-\i+.5}){$\i$}
}     
\foreach \iA in {1,...,\N}{\foreach \iB in {1,...,\N}{
    \edef\x{\calc{abs(\iA-\iB)}} 
    \ifnum\x=5
      \colorlet{fill color}{blue!20}
    \else
      \colorlet{fill color}{white}
    \fi
    \psframe[fillcolor=fill color](\iA, \calc{\N-\iB})(\calc{\iA+1}, \calc{\N-\iB+1})
    \rput(\calc{\iA+.5}, \calc{\N-\iB+.5}){$\calc{\iA*\iB}$} 
}}
\end{pspicture}    
\end{document}

在此处输入图片描述

Question 2

这似乎有点过度（事实也确实如此），但您具有很高的可定制性。

矩阵的自动构建是通过两个循环完成的，这两个循环在一个宏中添加正确数量的\pgfmatrixnextcells 和\\s，该宏仅提供给matrix内容。可能有更好的方法，但这是我目前发现的最简单的方法。

\pgfmathloop据我所知，LaTeX\loop带有一个计数器（\pgfmathcounter，一个宏），没有分组，也没有像 PGF 自己的那样的任何其他功能\foreach，但对于这一点来说已经足够了。

还有更多的改进空间，比如\pgfmatrix…从外部隐藏内部内容，例如以单独的方式与标题交互，让第一行成为实际的第一行包含内容等。

我添加了第二个图表，该图表使用函数 row ^columns来计算单元格的内容。第二列（即标题列之后的第一列）使用样式进行更改

column 2/.style={coordinate, column sep=+0pt},

这会将改column sep回零（-\pgflinewidth需要 sep，因为您希望线条重叠）并应用coordinate（即shape=coordinate）到节点，从而消除节点（及其文本）。当然，TikZ 仍然会创建一个命名坐标，所有...

这里针对列/行的进一步功能将会很有帮助。

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{matrix}
\tikzset{
  if/.code n args=3{\pgfmathparse{#1}\ifnum\pgfmathresult=0
    \pgfkeysalso{#3}\else\pgfkeysalso{#2}\fi},
  set column text width/.style={
    /utils/exec=\pgfmathtruncatemacro\multiTableTempa{#1-1}%
                \pgfmathtruncatemacro\multiTableTempb{multiTableFunction(\multiTableRows,\multiTableTempa)},
    column #1/.append style/.expanded={every node/.append style={text width=width("\multiTableTempb")}}},
  multiplication table/.style={
    matrix of math nodes, nodes in empty cells,
    column sep=+-\pgflinewidth, row sep=+-\pgflinewidth,
    cells={nodes={outer sep=+0pt,
      if={(\pgfmatrixcurrentrow-1)*(\pgfmatrixcurrentcolumn-1)==0}{}{multiTable every node/.try}}},
    row 1/.style={nodes={
      if={\pgfmatrixcurrentcolumn==1}{}{node contents=\pgfmathprint{int(\pgfmatrixcurrentcolumn-1)}},
      multiTable first row/.try}},
    column 1/.style={nodes={
      if={\pgfmatrixcurrentrow==1}{}{node contents=\pgfmathprint{int(\pgfmatrixcurrentrow-1)}},
      multiTable first column/.try}},
    row 1 column 1/.append style={nodes={align=center,
      node contents={\cdot},multiTable tr/.try}},
    every cell/.append style={
      nodes={
        minimum size=+1.5em, draw, align=right,
        text height=+.7\baselineskip, text depth=+0\baselineskip,
        if={and(\pgfmatrixcurrentrow>1,\pgfmatrixcurrentcolumn>1)}{
          node contents={\pgfmathprint{multiTableFunction(\pgfmatrixcurrentrow-1,\pgfmatrixcurrentcolumn-1)}},
        }{}}}},
  multiTable first row/.style={fill=orange!50},
  multiTable first column/.style={fill=orange!50},
  multiTable tr/.style={fill=orange!20},
  multiTable def function/.style={multiTable set function={int(\r*\c)}},
  multiTable set function/.style={declare function={multiTableFunction(\r,\c)=#1;}}}
\newcommand*\addTo[2]{\expandafter\def\expandafter#1\expandafter{#1#2}}
\newcommand*\multiplicationTable[3][]{%
\begin{tikzpicture}[#1]
\pgfmathtruncatemacro\multiTableRows{#2}
\pgfmathtruncatemacro\multiTableColumns{#3}
\def\multiTableTempa{}
\pgfmathloop
  \addTo\multiTableTempa{\pgfmatrixnextcell}
  \ifnum\pgfmathcounter<\multiTableColumns
\repeatpgfmathloop
\addTo\multiTableTempa{\\}\let\multiTableSetupFinal\multiTableTempa
\pgfmathloop
  \expandafter\addTo\expandafter\multiTableSetupFinal\expandafter{\multiTableTempa}
  \ifnum\pgfmathcounter<\multiTableRows
\repeatpgfmathloop
\pgfmathtruncatemacro\multiTableTempb{\multiTableColumns+1}
\matrix[multiplication table,
  set column text width/.list={2,...,\multiTableTempb},
  column 1/.append style/.expanded={nodes={text width=width("\multiTableRows")}},
]{\multiTableSetupFinal};
\end{tikzpicture}}
\begin{document}
\multiplicationTable[
  multiTable def function,
  every cell/.append style={
    if={(abs(\pgfmatrixcurrentrow-\pgfmatrixcurrentcolumn)==6)}{nodes={fill=red!50}}{}},
  ]{10}{10}
\multiplicationTable[
  multiTable set function={int(\r^\c)},
  multiTable tr/.append style={node contents=r^c,fill=blue!20},
  column 2/.style={coordinate, column sep=+0pt},
  multiTable first row/.style={fill=blue!50},
  multiTable first column/.style={fill=blue!50},
  multiTable every node/.style={text=blue},% blue enough?
  ]{10}{4}
\end{document}

输出

在此处输入图片描述

Answer

这似乎有点过度（事实也确实如此），但您具有很高的可定制性。

矩阵的自动构建是通过两个循环完成的，这两个循环在一个宏中添加正确数量的\pgfmatrixnextcells 和\\s，该宏仅提供给matrix内容。可能有更好的方法，但这是我目前发现的最简单的方法。

\pgfmathloop据我所知，LaTeX\loop带有一个计数器（\pgfmathcounter，一个宏），没有分组，也没有像 PGF 自己的那样的任何其他功能\foreach，但对于这一点来说已经足够了。

还有更多的改进空间，比如\pgfmatrix…从外部隐藏内部内容，例如以单独的方式与标题交互，让第一行成为实际的第一行包含内容等。

我添加了第二个图表，该图表使用函数 row ^columns来计算单元格的内容。第二列（即标题列之后的第一列）使用样式进行更改

column 2/.style={coordinate, column sep=+0pt},

这会将改column sep回零（-\pgflinewidth需要 sep，因为您希望线条重叠）并应用coordinate（即shape=coordinate）到节点，从而消除节点（及其文本）。当然，TikZ 仍然会创建一个命名坐标，所有...

这里针对列/行的进一步功能将会很有帮助。

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{matrix}
\tikzset{
  if/.code n args=3{\pgfmathparse{#1}\ifnum\pgfmathresult=0
    \pgfkeysalso{#3}\else\pgfkeysalso{#2}\fi},
  set column text width/.style={
    /utils/exec=\pgfmathtruncatemacro\multiTableTempa{#1-1}%
                \pgfmathtruncatemacro\multiTableTempb{multiTableFunction(\multiTableRows,\multiTableTempa)},
    column #1/.append style/.expanded={every node/.append style={text width=width("\multiTableTempb")}}},
  multiplication table/.style={
    matrix of math nodes, nodes in empty cells,
    column sep=+-\pgflinewidth, row sep=+-\pgflinewidth,
    cells={nodes={outer sep=+0pt,
      if={(\pgfmatrixcurrentrow-1)*(\pgfmatrixcurrentcolumn-1)==0}{}{multiTable every node/.try}}},
    row 1/.style={nodes={
      if={\pgfmatrixcurrentcolumn==1}{}{node contents=\pgfmathprint{int(\pgfmatrixcurrentcolumn-1)}},
      multiTable first row/.try}},
    column 1/.style={nodes={
      if={\pgfmatrixcurrentrow==1}{}{node contents=\pgfmathprint{int(\pgfmatrixcurrentrow-1)}},
      multiTable first column/.try}},
    row 1 column 1/.append style={nodes={align=center,
      node contents={\cdot},multiTable tr/.try}},
    every cell/.append style={
      nodes={
        minimum size=+1.5em, draw, align=right,
        text height=+.7\baselineskip, text depth=+0\baselineskip,
        if={and(\pgfmatrixcurrentrow>1,\pgfmatrixcurrentcolumn>1)}{
          node contents={\pgfmathprint{multiTableFunction(\pgfmatrixcurrentrow-1,\pgfmatrixcurrentcolumn-1)}},
        }{}}}},
  multiTable first row/.style={fill=orange!50},
  multiTable first column/.style={fill=orange!50},
  multiTable tr/.style={fill=orange!20},
  multiTable def function/.style={multiTable set function={int(\r*\c)}},
  multiTable set function/.style={declare function={multiTableFunction(\r,\c)=#1;}}}
\newcommand*\addTo[2]{\expandafter\def\expandafter#1\expandafter{#1#2}}
\newcommand*\multiplicationTable[3][]{%
\begin{tikzpicture}[#1]
\pgfmathtruncatemacro\multiTableRows{#2}
\pgfmathtruncatemacro\multiTableColumns{#3}
\def\multiTableTempa{}
\pgfmathloop
  \addTo\multiTableTempa{\pgfmatrixnextcell}
  \ifnum\pgfmathcounter<\multiTableColumns
\repeatpgfmathloop
\addTo\multiTableTempa{\\}\let\multiTableSetupFinal\multiTableTempa
\pgfmathloop
  \expandafter\addTo\expandafter\multiTableSetupFinal\expandafter{\multiTableTempa}
  \ifnum\pgfmathcounter<\multiTableRows
\repeatpgfmathloop
\pgfmathtruncatemacro\multiTableTempb{\multiTableColumns+1}
\matrix[multiplication table,
  set column text width/.list={2,...,\multiTableTempb},
  column 1/.append style/.expanded={nodes={text width=width("\multiTableRows")}},
]{\multiTableSetupFinal};
\end{tikzpicture}}
\begin{document}
\multiplicationTable[
  multiTable def function,
  every cell/.append style={
    if={(abs(\pgfmatrixcurrentrow-\pgfmatrixcurrentcolumn)==6)}{nodes={fill=red!50}}{}},
  ]{10}{10}
\multiplicationTable[
  multiTable set function={int(\r^\c)},
  multiTable tr/.append style={node contents=r^c,fill=blue!20},
  column 2/.style={coordinate, column sep=+0pt},
  multiTable first row/.style={fill=blue!50},
  multiTable first column/.style={fill=blue!50},
  multiTable every node/.style={text=blue},% blue enough?
  ]{10}{4}
\end{document}

输出

在此处输入图片描述

Question 3

如果需要 TikZ，

\documentclass[tikz]{standalone}
\usetikzlibrary{matrix}
\def\mcr{\pgfmatrixcurrentrow}\def\mcc{\pgfmatrixcurrentcolumn}
\begin{document}
\begin{tikzpicture}[border style/.style={
    draw,fill=#1,minimum size=0.8cm,anchor=center,outer sep=0,
    name=\tikzmatrixname-\the\mcr-\the\mcc
}]
\matrix[row sep=-.5*\pgflinewidth,column sep=-.5*\pgflinewidth,
   execute at empty cell={%
       \ifnum1=\mcr\relax%
           \ifnum1=\mcc\relax\node[border style=red]{$\cdot$};%
           \else\node[border style=red]{$\number\numexpr\the\mcc-1\relax$};\fi
       \else%
         \ifnum1=\mcc\relax\node[border style=red]{$\cdot$};%
           \node[border style=red]{$\number\numexpr\the\mcr-1\relax$};%
         \else%
           \pgfmathparse{int(abs(\mcr-\mcc))}%
           \ifnum5=\pgfmathresult\relax\def\temp{yellow}\else\def\temp{none}\fi%
           \node[border style=\temp]{\number\numexpr\numexpr\the\mcc-1\relax*\numexpr\the\mcr-1\relax\relax};%
         \fi%
       \fi}
] (a) {
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
};
\end{tikzpicture}
\end{document}

在此处输入图片描述

Answer

如果需要 TikZ，

\documentclass[tikz]{standalone}
\usetikzlibrary{matrix}
\def\mcr{\pgfmatrixcurrentrow}\def\mcc{\pgfmatrixcurrentcolumn}
\begin{document}
\begin{tikzpicture}[border style/.style={
    draw,fill=#1,minimum size=0.8cm,anchor=center,outer sep=0,
    name=\tikzmatrixname-\the\mcr-\the\mcc
}]
\matrix[row sep=-.5*\pgflinewidth,column sep=-.5*\pgflinewidth,
   execute at empty cell={%
       \ifnum1=\mcr\relax%
           \ifnum1=\mcc\relax\node[border style=red]{$\cdot$};%
           \else\node[border style=red]{$\number\numexpr\the\mcc-1\relax$};\fi
       \else%
         \ifnum1=\mcc\relax\node[border style=red]{$\cdot$};%
           \node[border style=red]{$\number\numexpr\the\mcr-1\relax$};%
         \else%
           \pgfmathparse{int(abs(\mcr-\mcc))}%
           \ifnum5=\pgfmathresult\relax\def\temp{yellow}\else\def\temp{none}\fi%
           \node[border style=\temp]{\number\numexpr\numexpr\the\mcc-1\relax*\numexpr\the\mcr-1\relax\relax};%
         \fi%
       \fi}
] (a) {
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
&&&&&&&&&&&\\
};
\end{tikzpicture}
\end{document}

在此处输入图片描述

Question 4

MetaPost 是另一个适合此类图片的候选者。

input mpcolornames; color myorange; myorange = .5[Red,Yellow];
input latexmp; setupLaTeXMP(mode = rerun, textextlabel = enable);
u := 1cm;
def loc(expr i,j) = 
  ((i-.5, -.5-j) -- (i+.5, -.5-j) -- (i+.5, .5-j) -- (i-.5, .5-j) -- cycle) scaled u
enddef;
beginfig(1);
  fill loc(0, 0) withcolor myorange; label(".", origin); 
  for i = 1 upto 11:
    fill loc(i, 0) withcolor myorange;
    label(decimal i, (i*u, 0));
    fill loc(0, i) withcolor myorange;
    label(decimal i, (0, -i*u));
  endfor
  % Filling
  for i = 1 upto 11:
    for j = 1 upto 11:
      if abs(i-j) = 5: fill loc(i,j) withcolor Yellow; fi
      label(decimal(i*j), u*(i, -j));
    endfor
  endfor
  % Grid lines
  for i = 0 upto 12:
    draw u*(i-0.5, .5) -- u*(i-0.5, -11.5);
    draw u*(-0.5, 0.5-i) -- u*(11.5, 0.5-i);
  endfor;
endfig;
end.

对于每对(i,j)整数，宏loc(i,j)都会在表中给出其位置（i*j按此顺序包含的方块），并且如果，则以下指令用黄色填充该位置|i-j|=5：

if abs(i-j) = 5: fill loc(i,j) withcolor Yellow; fi

结果：

在此处输入图片描述

Answer

MetaPost 是另一个适合此类图片的候选者。

input mpcolornames; color myorange; myorange = .5[Red,Yellow];
input latexmp; setupLaTeXMP(mode = rerun, textextlabel = enable);
u := 1cm;
def loc(expr i,j) = 
  ((i-.5, -.5-j) -- (i+.5, -.5-j) -- (i+.5, .5-j) -- (i-.5, .5-j) -- cycle) scaled u
enddef;
beginfig(1);
  fill loc(0, 0) withcolor myorange; label(".", origin); 
  for i = 1 upto 11:
    fill loc(i, 0) withcolor myorange;
    label(decimal i, (i*u, 0));
    fill loc(0, i) withcolor myorange;
    label(decimal i, (0, -i*u));
  endfor
  % Filling
  for i = 1 upto 11:
    for j = 1 upto 11:
      if abs(i-j) = 5: fill loc(i,j) withcolor Yellow; fi
      label(decimal(i*j), u*(i, -j));
    endfor
  endfor
  % Grid lines
  for i = 0 upto 12:
    draw u*(i-0.5, .5) -- u*(i-0.5, -11.5);
    draw u*(-0.5, 0.5-i) -- u*(11.5, 0.5-i);
  endfor;
endfig;
end.

对于每对(i,j)整数，宏loc(i,j)都会在表中给出其位置（i*j按此顺序包含的方块），并且如果，则以下指令用黄色填充该位置|i-j|=5：

if abs(i-j) = 5: fill loc(i,j) withcolor Yellow; fi

结果：

在此处输入图片描述

自动生成乘法表并指定颜色的行、列和对角线

答案1

答案2

代码

输出

答案3

答案4

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