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如何突出显示帕斯卡三角形中的四面体数

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如何突出显示帕斯卡三角形中的四面体数

四面体数表示创建具有 n 层的三角金字塔所需的半径相等的球的数量。第 n 个四面体数由公式 C(n + 2, 3) 给出。因此，四面体数在帕斯卡三角形中显示为对角线。我希望在帕斯卡三角形中突出显示这些数字。但是，我的尝试产生的是帕斯卡三角形，其中突出显示的数字单独显示为红色。我如何修改代码，以便四面体数在三角形本身中突出显示？

四面体数

以下是代码，基于 Caramdir 对这个问题关于如何以数字形式表示帕斯卡三角形中的二项式系数：

\documentclass{article}
\usepackage{tikz}

%calculate binomial coefficients 
\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}

\begin{figure}[h]
\centering
\begin{tikzpicture}
        \foreach \n in {0, ..., 7} {
            \foreach \k in {0,...,\n} {
                \node (\n\k) at (\k-\n/2,-\n) {\(\binomialcoefficient{\n}{\k}\)};
                \foreach \n in {3, 4, ..., 7} \node[color = red] at (\n, 3) {\(\binomialcoefficient{\n}{3}\)};
            }
            \pgfmathtruncatemacro{\x}{(\n+1)/2}
            \pgfmathtruncatemacro{\y}{\n/2}
        }
\end{tikzpicture}
\end{figure}

\end{document}

答案1

\n您的算法中的四面体数是通过指标与指标之间的差等于来识别\k的3。

因此，只需计算这个差异并按如下方式着色：

\pgfmathparse{int(\n-\k)}
\ifnum \pgfmathresult=3
\node[red,node font=\bf] (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
\else
\node (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
\fi

截屏

\documentclass{article}
\usepackage{tikz}

%calculate binomial coefficients 
\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}

\begin{figure}[h]
\centering
\begin{tikzpicture}
        \foreach \n in {0, ...,7} {
            \foreach \k in {0,...,\n} {
             \pgfmathparse{int(\n-\k)}
             \ifnum \pgfmathresult=3
             \node[red,node font=\bf] (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
             \else
             \node (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
             \fi
%                \foreach \n in {3, 4, ..., 7} \node[color = red] at (\n, 3) {\(\binomialcoefficient{\n}{3}\)};
            }
            \pgfmathtruncatemacro{\x}{(\n+1)/2}
            \pgfmathtruncatemacro{\y}{\n/2}
        }
\end{tikzpicture}
\end{figure}

\end{document}

答案2

AndreC 对我来说太快了 :) 只是一种直接评估的替代方法\x，\y直接\n-\k在循环中评估并避免\pgfmathtruncatemacro

\documentclass{standalone}
\usepackage{tikz}

%calculate binomial coefficients 
\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}

\begin{tikzpicture}
    \foreach \n in {0, ..., 11} 
    {
        \foreach [evaluate ={
                \x = int(0.5*(\n+1));
                \y = int(0.5*\n) ;
                \NmK = int(\n-\k) ; %n minus k, NmK
            }] \k in {0,...,\n} 
        {
            \ifnum  \NmK=3          
                \node[red,node font=\bf] (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
            \else
                \node (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
            \fi             

        }
    }
\end{tikzpicture}

\end{document}

答案3

AndreC 和 JeT 提出的优秀解决方案都突出显示了 C(n + 2, n - 1) 项，其中 n 至少为 1。由于我感兴趣的是突出显示 C(n + 2, 3) 项，同样 n 至少为 1，因此我修改了 AndreC 的代码以突出显示对角线，如下所示：

\documentclass{article}
\usepackage{tikz}

%calculate binomial coefficients 
\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}

\begin{figure}[h]
\centering
\begin{tikzpicture}
        \foreach \n in {0, ...,7} {
            \foreach \k in {0,...,\n} {
             \ifnum \k=3
             \node[red,node font=\bf] (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
             \else
             \node (\n\k) at (\k-\n/2,-\n) {\binomialcoefficient{\n}{\k}};
             \fi
%                \foreach \n in {3, 4, ..., 7} \node[color = red] at (\n, 3) {\(\binomialcoefficient{\n}{3}\)};
            }
            \pgfmathtruncatemacro{\x}{(\n+1)/2}
            \pgfmathtruncatemacro{\y}{\n/2}
        }
\end{tikzpicture}

\end{figure}

\end{document}

帕斯卡三角形与四面体数字突出显示

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