带有空行的六边形帕斯卡三角形

Question 1

您需要根据以下值更改节点文本\newn：

定义最后一行非空行：

\def\BlankRowAfter{6}

进而

\ifnum\newn>\BlankRowAfter\relax
  \def\NodeText{}%
\else
  \def\NodeText{\tiny $\mathbf{\binomialCoefficient{\newn}{\k}}$}%
\fi

如果需要则确定\NodeText为空白。

代码：

\documentclass{article}%
\usepackage[dvipsnames]{xcolor} \pagecolor{white}
\usepackage{tikz}
\usepackage{ifthen}
\makeatletter
\newcommand\binomialCoefficient[2]{%
    % Store values 
    \c@pgf@counta=#1% n
    \c@pgf@countb=#2% k
    %
    % Take advantage of symmetry if k > n - k
    \c@pgf@countc=\c@pgf@counta%
    \advance\c@pgf@countc by-\c@pgf@countb%
    \ifnum\c@pgf@countb>\c@pgf@countc%
        \c@pgf@countb=\c@pgf@countc%
    \fi%
    %
    % Recursively compute the coefficients
    \c@pgf@countc=1% will hold the result
    \c@pgf@countd=0% counter
    \pgfmathloop% c -> c*(n-i)/(i+1) for i=0,...,k-1
        \ifnum\c@pgf@countd<\c@pgf@countb%
        \multiply\c@pgf@countc by\c@pgf@counta%
        \advance\c@pgf@counta by-1%
        \advance\c@pgf@countd by1%
        \divide\c@pgf@countc by\c@pgf@countd%
    \repeatpgfmathloop%
    \the\c@pgf@countc%
}
\makeatother
\begin{document}
\newdimen\R
\R=.4cm
\newcommand\mycolor{white}
\begin{center}
    \begin{tikzpicture}[line width=.8pt]
        \def\rows{9}
        \def\BlankRowAfter{6}
        \foreach \k in {0,...,\rows}
        {
            \begin{scope}[shift={(-60:{sqrt(3)*\R*\k})}]
                \pgfmathtruncatemacro\ystart{\rows-\k}
                \foreach \n in {0,...,\ystart}
                {
                    \pgfmathtruncatemacro\newn{\n+\k}
                    % \ifthenelse{\k=0}{\def\mycolor{pink}}{}
                    % \ifthenelse{\k=1}{\def\mycolor{yellow}}{}
                    % \ifthenelse{\k=2}{\def\mycolor{blue}}{}
                    % \ifthenelse{\k=3}{\def\mycolor{green}}{}
                    % \ifthenelse{\k=8 \AND \n < 4}{\def\mycolor{red}}{}
                    % \ifthenelse{\k=9 \AND \n = 3}{\def\mycolor{purple}}{}
                    \ifnum\newn>\BlankRowAfter\relax
                        \def\NodeText{}%
                    \else
                        \def\NodeText{\tiny $\mathbf{\binomialCoefficient{\newn}{\k}}$}%
                    \fi
                    \begin{scope}[shift={(-120:{sqrt(3)*\R*\n})}]
                        \draw[top color=\mycolor!20,bottom color=\mycolor!60] 
                         (30:\R) \foreach \x in {90,150,...,330} {
                            -- (\x:\R)}
                            -- cycle (90:0)
                               node {\NodeText};
                    \end{scope}
                }
            \end{scope}
        }
    \end{tikzpicture} 
\end{center}
\end{document}

Answer

您需要根据以下值更改节点文本\newn：

定义最后一行非空行：

\def\BlankRowAfter{6}

进而

\ifnum\newn>\BlankRowAfter\relax
  \def\NodeText{}%
\else
  \def\NodeText{\tiny $\mathbf{\binomialCoefficient{\newn}{\k}}$}%
\fi

如果需要则确定\NodeText为空白。

代码：

\documentclass{article}%
\usepackage[dvipsnames]{xcolor} \pagecolor{white}
\usepackage{tikz}
\usepackage{ifthen}
\makeatletter
\newcommand\binomialCoefficient[2]{%
    % Store values 
    \c@pgf@counta=#1% n
    \c@pgf@countb=#2% k
    %
    % Take advantage of symmetry if k > n - k
    \c@pgf@countc=\c@pgf@counta%
    \advance\c@pgf@countc by-\c@pgf@countb%
    \ifnum\c@pgf@countb>\c@pgf@countc%
        \c@pgf@countb=\c@pgf@countc%
    \fi%
    %
    % Recursively compute the coefficients
    \c@pgf@countc=1% will hold the result
    \c@pgf@countd=0% counter
    \pgfmathloop% c -> c*(n-i)/(i+1) for i=0,...,k-1
        \ifnum\c@pgf@countd<\c@pgf@countb%
        \multiply\c@pgf@countc by\c@pgf@counta%
        \advance\c@pgf@counta by-1%
        \advance\c@pgf@countd by1%
        \divide\c@pgf@countc by\c@pgf@countd%
    \repeatpgfmathloop%
    \the\c@pgf@countc%
}
\makeatother
\begin{document}
\newdimen\R
\R=.4cm
\newcommand\mycolor{white}
\begin{center}
    \begin{tikzpicture}[line width=.8pt]
        \def\rows{9}
        \def\BlankRowAfter{6}
        \foreach \k in {0,...,\rows}
        {
            \begin{scope}[shift={(-60:{sqrt(3)*\R*\k})}]
                \pgfmathtruncatemacro\ystart{\rows-\k}
                \foreach \n in {0,...,\ystart}
                {
                    \pgfmathtruncatemacro\newn{\n+\k}
                    % \ifthenelse{\k=0}{\def\mycolor{pink}}{}
                    % \ifthenelse{\k=1}{\def\mycolor{yellow}}{}
                    % \ifthenelse{\k=2}{\def\mycolor{blue}}{}
                    % \ifthenelse{\k=3}{\def\mycolor{green}}{}
                    % \ifthenelse{\k=8 \AND \n < 4}{\def\mycolor{red}}{}
                    % \ifthenelse{\k=9 \AND \n = 3}{\def\mycolor{purple}}{}
                    \ifnum\newn>\BlankRowAfter\relax
                        \def\NodeText{}%
                    \else
                        \def\NodeText{\tiny $\mathbf{\binomialCoefficient{\newn}{\k}}$}%
                    \fi
                    \begin{scope}[shift={(-120:{sqrt(3)*\R*\n})}]
                        \draw[top color=\mycolor!20,bottom color=\mycolor!60] 
                         (30:\R) \foreach \x in {90,150,...,330} {
                            -- (\x:\R)}
                            -- cycle (90:0)
                               node {\NodeText};
                    \end{scope}
                }
            \end{scope}
        }
    \end{tikzpicture} 
\end{center}
\end{document}

Question 2

我喜欢 TikZ 自带的按键系统。

在下面的代码中，针对路径分别尝试了三种样式：

every binom path = {k}{n}以及 binom path row nbinom path col k
every binom node = {k}{n}，和。binom node row nbinom node col k

这样，您就可以定义特定行或列的样式，也可以通过在样式中测试#1和来为每个元素定义样式。在这种情况下，我们设置，和，以便将设置为空，并且其他人无法通过禁用键再次重置它。#2every binom ....binom node row 7… 8… 9node contentsnode contents

我还添加了二项式系数的可扩展 eTeX 评估。

我们还regular polygon可以使用节点来绘制六边形，但这并不会使其变得更容易。

代码

\documentclass[tikz]{standalone}\usepackage{libertine}
\newcommand*\binomialCoefficient[2]{% #1 = n, #2 = k
  \expanded{\noexpand\binomialCoefficientRec{\pgfintmin{#1-#2}{#2}}}{1}{#1}{1}}
\newcommand*\binomialCoefficientRec[4]{% #1 = k, #2 = i, #3 = n, #4 = result
  \ifnum#2>#1 #4\else
    \expanded{\noexpand\binomialCoefficientRec{#1}{\pgfinteval{#2+1}}{#3}
      {\pgfintdivtruncate{#4*(#3+1-(#2))}{#2}}}\fi}
\begin{document}
\tikz[
  thick, no text/.style={node contents=, node contents/.code=},
  every binom node/.style 2 args={node contents={$\binomialCoefficient{#2}{#1}$}},
  no binom text/.style={binom node row #1/.style={no text}},
  no binom text/.list={7, 8, 9}
]
\foreach \n in {0,..., 9}
  \foreach \k in {0, ..., \n}
    \draw[
      shift={([y=(-120:1)]\k,\n)}, every binom path/.try={\k}{\n},
      binom path row \n/.try, binom path col \k/.try]
        node[node contents=, every binom node/.try={\k}{\n},
             binom node row \n/.try, binom node col \k/.try]
        plot[sharp cycle, samples at={0, ..., 5}](30+\x*60:sqrt 3/3);
\end{document}

输出

Answer