[Tex/LaTex] Pascal’s triangle in tikz with row headers

loopstikz-pgf

I am interested in creating Pascal's triangle with Tikz and LaTeX. I would like to add a column of text on the left of the triangle, with header "Row Number" and then the entries of each row. I am new to Tikz and had trouble creating a structure like this on my own.

Like this (center justified triangle, not left aligned)

Row 0: 1
Row 1: 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 56 28 8 1
Row 9: 1 9 36 84 126 126 84 36 9 1
Row 10: 1 10 45 120 210 252 210 120 45 10 1

This code generated ideas, though since I am new to tikz, I wasn't sure where to go with it.

Pascal's triangle in tikz

Thanks!

Best Answer

The following should do it. By using the fpu library it is possible to go beyond the usual limit of 17 rows in PGF.

\documentclass[tikz,border=5]{standalone}
\usepgflibrary{fpu}
\begin{document}
\def\N{10}
\tikz[x=0.75cm,y=0.5cm, 
  pascal node/.style={font=\footnotesize}, 
  row node/.style={font=\footnotesize, anchor=west, shift=(180:1)}]
  \path  
    \foreach \n in {0,...,\N} { 
      (-\N/2-1, -\n) node  [row node/.try]{Row \n:}
        \foreach \k in {0,...,\n}{
          (-\n/2+\k,-\n) node [pascal node/.try] {%
            \pgfkeys{/pgf/fpu}%
            \pgfmathparse{round(\n!/(\k!*(\n-\k)!))}%
            \pgfmathfloattoint{\pgfmathresult}%
            \pgfmathresult%
          }}};
\end{document}

enter image description here

Related Solutions

[Tex/LaTex] Pascal’s triangle in tikz

Here is a solution using TeX integer arithmetic. I am reusing counters defined by PGF in order to avoid having to declare new ones.

\documentclass{article}
\usepackage{tikz}

\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,...,15} {
  \foreach \k in {0,...,\n} {
    \node at (\k-\n/2,-\n) {$\binomialCoefficient{\n}{\k}$};
  }
}
\end{tikzpicture}

\end{document}

enter image description here

If you want, you can wrap \pgfmathdeclarefunction around that to have the function available in pgfmath (see Section 65 “Customizing the Mathematical Engine” in the manual (v2.10)).

[Tex/LaTex] How to get access to the size parameters of a TikZ node

Here is one possibility with not-exactly-failproof node construction. The triangles are also accesible with (<nodename>-leftri) and (<nodename>-rightri). The possible failure reason would be rotating the node. It's possible to include that but it's unnecessarily complicated at this stage and I don't know if you ever need it.

\documentclass[tikz]{standalone}
\usetikzlibrary{shapes.geometric}

\begin{document}

\begin{tikzpicture}[mynode/.style args={#1 and #2}{
draw,
minimum height=#1,
minimum width=#2,
rounded corners=0.5*#1,
append after command={
\pgfextra{%                 
    \begin{pgfinterruptpath}
    \node[isosceles triangle,
          isosceles triangle stretches,
          draw=black,inner sep=0,
          anchor=west,
          minimum width=0.5*#1,
          minimum height=0.2*#1,
          fill=black] 
          (\tikzlastnode-leftri) at ([xshift=0.5*#1]\tikzlastnode.west) {};
    \node[isosceles triangle,
          isosceles triangle stretches,
          draw=black,inner sep=0,
          anchor=west,
          minimum width=0.5*#1,
          minimum height=0.2*#1,
          isosceles triangle apex angle=90]
          (\tikzlastnode-rightri) at ([xshift=-0.5*#1]\tikzlastnode.east) {};
\end{pgfinterruptpath}
        }   
    }
}
]

\node[mynode=1cm and 4cm] (a) {};
\node[mynode=3cm and 8cm] (b) at (0,3) {};
\draw (a-leftri) -- (b-rightri);
\end{tikzpicture}

\end{document}

enter image description here

Best Answer

Related Solutions

[Tex/LaTex] Pascal’s triangle in tikz

[Tex/LaTex] How to get access to the size parameters of a TikZ node

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