How can I make a graph rho-shaped whose nodes are numbers?












6














I managed to get the two pictures below. I think I'd have to rotate the straight line and connect its end to the number 615 so that it looks like the letter rho (ρ). How can I do that? Feel free to suggest a whole different approach.



enter image description here



Notice how the arrows of the straight line are too small compared to the ones in the cycle. I'd also appreciate seeing an approach where these arrow sizes are similar. It's okay if I have to type up each node — as long defining their position is a sane procedure. I'm sure whatever you come up with will better than I could do today — having read only the first pages of the TikZ manual.



% A simple cycle
% Author : Jerome Tremblay
documentclass{article}
usepackage{tikz}
begin{document}

begin{tikzpicture}
def n {11}
def radius {3.5cm}
def margin {8} % margin in angles, depends on the radius

%% draw[step=.5cm,gray,very thin] (-7,-7) grid (7,7);
%% draw (0,0) circle (0.1cm);
%% --8<---------------cut here---------------start------------->8---
%% the tail
%% path
%% (0,0) node [shape=circle,draw,minimum width=1cm]{$2$}
%% (1,0) node [shape=circle,draw,minimum width=1cm]{$12$}
%% (2,0) node [shape=circle,draw,minimum width=1cm]{$152$}
%% (3,0) node [shape=circle,draw,minimum width=1cm]{$1223$}
%% (4,0) node [shape=circle,draw,minimum width=1cm]{$1031$}
%% (5,0) node [shape=circle,draw,minimum width=1cm]{$2916$}
%% (6,0) node [shape=circle,draw,minimum width=1cm]{$751$}
%% (7,0) node [shape=circle,draw,minimum width=1cm]{$1149$};


%% --8<---------------cut here---------------end--------------->8---

% the cycle
def s {1}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$456$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {2}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$1562$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {3}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$792$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {4}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$1872$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {5}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$2152$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {6}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$25$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {7}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$441$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {8}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$615$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {9}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$2993$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {10}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$2329$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

def s {11}
node[draw=none] at ({360/n * -(s - 1)}:radius) {$2031$};
draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

end{tikzpicture}

begin{tikzpicture}[node distance=1.5cm,minimum width=1.2cm]
node(n0) [draw=none] at (-15,0) {$2$};
node (n1) [draw=none,right of=n0]{$12$};
node (n2) [draw=none,right of=n1]{$152$};
node (n3) [draw=none,right of=n2]{$1223$};
node (n4) [draw=none,right of=n3]{$1031$};
node (n5) [draw=none,right of=n4]{$2916$};
node (n6) [draw=none,right of=n5]{$751$};
node (n7) [draw=none,right of=n6]{$1149$};
draw [->] (n0) to (n1);
draw [->] (n1) to (n2);
draw [->] (n2) to (n3);
draw [->] (n3) to (n4);
draw [->] (n4) to (n5);
draw [->] (n5) to (n6);
draw [->] (n6) to (n7);
end{tikzpicture}

end{document}









share|improve this question





























    6














    I managed to get the two pictures below. I think I'd have to rotate the straight line and connect its end to the number 615 so that it looks like the letter rho (ρ). How can I do that? Feel free to suggest a whole different approach.



    enter image description here



    Notice how the arrows of the straight line are too small compared to the ones in the cycle. I'd also appreciate seeing an approach where these arrow sizes are similar. It's okay if I have to type up each node — as long defining their position is a sane procedure. I'm sure whatever you come up with will better than I could do today — having read only the first pages of the TikZ manual.



    % A simple cycle
    % Author : Jerome Tremblay
    documentclass{article}
    usepackage{tikz}
    begin{document}

    begin{tikzpicture}
    def n {11}
    def radius {3.5cm}
    def margin {8} % margin in angles, depends on the radius

    %% draw[step=.5cm,gray,very thin] (-7,-7) grid (7,7);
    %% draw (0,0) circle (0.1cm);
    %% --8<---------------cut here---------------start------------->8---
    %% the tail
    %% path
    %% (0,0) node [shape=circle,draw,minimum width=1cm]{$2$}
    %% (1,0) node [shape=circle,draw,minimum width=1cm]{$12$}
    %% (2,0) node [shape=circle,draw,minimum width=1cm]{$152$}
    %% (3,0) node [shape=circle,draw,minimum width=1cm]{$1223$}
    %% (4,0) node [shape=circle,draw,minimum width=1cm]{$1031$}
    %% (5,0) node [shape=circle,draw,minimum width=1cm]{$2916$}
    %% (6,0) node [shape=circle,draw,minimum width=1cm]{$751$}
    %% (7,0) node [shape=circle,draw,minimum width=1cm]{$1149$};


    %% --8<---------------cut here---------------end--------------->8---

    % the cycle
    def s {1}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$456$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {2}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$1562$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {3}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$792$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {4}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$1872$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {5}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$2152$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {6}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$25$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {7}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$441$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {8}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$615$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {9}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$2993$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {10}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$2329$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    def s {11}
    node[draw=none] at ({360/n * -(s - 1)}:radius) {$2031$};
    draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
    arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

    end{tikzpicture}

    begin{tikzpicture}[node distance=1.5cm,minimum width=1.2cm]
    node(n0) [draw=none] at (-15,0) {$2$};
    node (n1) [draw=none,right of=n0]{$12$};
    node (n2) [draw=none,right of=n1]{$152$};
    node (n3) [draw=none,right of=n2]{$1223$};
    node (n4) [draw=none,right of=n3]{$1031$};
    node (n5) [draw=none,right of=n4]{$2916$};
    node (n6) [draw=none,right of=n5]{$751$};
    node (n7) [draw=none,right of=n6]{$1149$};
    draw [->] (n0) to (n1);
    draw [->] (n1) to (n2);
    draw [->] (n2) to (n3);
    draw [->] (n3) to (n4);
    draw [->] (n4) to (n5);
    draw [->] (n5) to (n6);
    draw [->] (n6) to (n7);
    end{tikzpicture}

    end{document}









    share|improve this question



























      6












      6








      6


      0





      I managed to get the two pictures below. I think I'd have to rotate the straight line and connect its end to the number 615 so that it looks like the letter rho (ρ). How can I do that? Feel free to suggest a whole different approach.



      enter image description here



      Notice how the arrows of the straight line are too small compared to the ones in the cycle. I'd also appreciate seeing an approach where these arrow sizes are similar. It's okay if I have to type up each node — as long defining their position is a sane procedure. I'm sure whatever you come up with will better than I could do today — having read only the first pages of the TikZ manual.



      % A simple cycle
      % Author : Jerome Tremblay
      documentclass{article}
      usepackage{tikz}
      begin{document}

      begin{tikzpicture}
      def n {11}
      def radius {3.5cm}
      def margin {8} % margin in angles, depends on the radius

      %% draw[step=.5cm,gray,very thin] (-7,-7) grid (7,7);
      %% draw (0,0) circle (0.1cm);
      %% --8<---------------cut here---------------start------------->8---
      %% the tail
      %% path
      %% (0,0) node [shape=circle,draw,minimum width=1cm]{$2$}
      %% (1,0) node [shape=circle,draw,minimum width=1cm]{$12$}
      %% (2,0) node [shape=circle,draw,minimum width=1cm]{$152$}
      %% (3,0) node [shape=circle,draw,minimum width=1cm]{$1223$}
      %% (4,0) node [shape=circle,draw,minimum width=1cm]{$1031$}
      %% (5,0) node [shape=circle,draw,minimum width=1cm]{$2916$}
      %% (6,0) node [shape=circle,draw,minimum width=1cm]{$751$}
      %% (7,0) node [shape=circle,draw,minimum width=1cm]{$1149$};


      %% --8<---------------cut here---------------end--------------->8---

      % the cycle
      def s {1}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$456$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {2}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$1562$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {3}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$792$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {4}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$1872$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {5}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2152$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {6}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$25$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {7}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$441$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {8}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$615$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {9}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2993$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {10}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2329$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {11}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2031$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      end{tikzpicture}

      begin{tikzpicture}[node distance=1.5cm,minimum width=1.2cm]
      node(n0) [draw=none] at (-15,0) {$2$};
      node (n1) [draw=none,right of=n0]{$12$};
      node (n2) [draw=none,right of=n1]{$152$};
      node (n3) [draw=none,right of=n2]{$1223$};
      node (n4) [draw=none,right of=n3]{$1031$};
      node (n5) [draw=none,right of=n4]{$2916$};
      node (n6) [draw=none,right of=n5]{$751$};
      node (n7) [draw=none,right of=n6]{$1149$};
      draw [->] (n0) to (n1);
      draw [->] (n1) to (n2);
      draw [->] (n2) to (n3);
      draw [->] (n3) to (n4);
      draw [->] (n4) to (n5);
      draw [->] (n5) to (n6);
      draw [->] (n6) to (n7);
      end{tikzpicture}

      end{document}









      share|improve this question















      I managed to get the two pictures below. I think I'd have to rotate the straight line and connect its end to the number 615 so that it looks like the letter rho (ρ). How can I do that? Feel free to suggest a whole different approach.



      enter image description here



      Notice how the arrows of the straight line are too small compared to the ones in the cycle. I'd also appreciate seeing an approach where these arrow sizes are similar. It's okay if I have to type up each node — as long defining their position is a sane procedure. I'm sure whatever you come up with will better than I could do today — having read only the first pages of the TikZ manual.



      % A simple cycle
      % Author : Jerome Tremblay
      documentclass{article}
      usepackage{tikz}
      begin{document}

      begin{tikzpicture}
      def n {11}
      def radius {3.5cm}
      def margin {8} % margin in angles, depends on the radius

      %% draw[step=.5cm,gray,very thin] (-7,-7) grid (7,7);
      %% draw (0,0) circle (0.1cm);
      %% --8<---------------cut here---------------start------------->8---
      %% the tail
      %% path
      %% (0,0) node [shape=circle,draw,minimum width=1cm]{$2$}
      %% (1,0) node [shape=circle,draw,minimum width=1cm]{$12$}
      %% (2,0) node [shape=circle,draw,minimum width=1cm]{$152$}
      %% (3,0) node [shape=circle,draw,minimum width=1cm]{$1223$}
      %% (4,0) node [shape=circle,draw,minimum width=1cm]{$1031$}
      %% (5,0) node [shape=circle,draw,minimum width=1cm]{$2916$}
      %% (6,0) node [shape=circle,draw,minimum width=1cm]{$751$}
      %% (7,0) node [shape=circle,draw,minimum width=1cm]{$1149$};


      %% --8<---------------cut here---------------end--------------->8---

      % the cycle
      def s {1}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$456$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {2}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$1562$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {3}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$792$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {4}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$1872$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {5}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2152$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {6}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$25$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {7}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$441$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {8}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$615$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {9}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2993$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {10}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2329$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      def s {11}
      node[draw=none] at ({360/n * -(s - 1)}:radius) {$2031$};
      draw[<-, >=latex] ({360/n * (s - 1)+margin}:radius)
      arc ({360/n * (s - 1)+margin}:{360/n * (s)-margin}:radius);

      end{tikzpicture}

      begin{tikzpicture}[node distance=1.5cm,minimum width=1.2cm]
      node(n0) [draw=none] at (-15,0) {$2$};
      node (n1) [draw=none,right of=n0]{$12$};
      node (n2) [draw=none,right of=n1]{$152$};
      node (n3) [draw=none,right of=n2]{$1223$};
      node (n4) [draw=none,right of=n3]{$1031$};
      node (n5) [draw=none,right of=n4]{$2916$};
      node (n6) [draw=none,right of=n5]{$751$};
      node (n7) [draw=none,right of=n6]{$1149$};
      draw [->] (n0) to (n1);
      draw [->] (n1) to (n2);
      draw [->] (n2) to (n3);
      draw [->] (n3) to (n4);
      draw [->] (n4) to (n5);
      draw [->] (n5) to (n6);
      draw [->] (n6) to (n7);
      end{tikzpicture}

      end{document}






      tikz-pgf






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      asked 2 days ago









      Joep AwinitaJoep Awinita

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          Your circle and straight lines can be coded much shorter, which also allows you to change the positions of the various nodes very easily.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain},
          every join/.style=-latex]]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here



          Just for fun: the straight line of a latex rho has an angle of approximately 76 degrees.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}
          node[scale=15]{$rho$};
          draw[white,thick,double=blue] (-0.5,0.6) -- (-1.05,-1.6)
          node[midway,left=5mm,scale=3,blue]{pgfmathparse{atan2(0.6+1.6,1.05-0.5)}%
          $pgfmathprintnumber{pgfmathresult}^circ$};
          end{tikzpicture}
          end{document}


          enter image description here



          This raises the question if one can make the chain such that it has this angle. The answer is yes.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain,
          ,xshift=-{cot(76)*1.5cm}},
          every join/.style=-latex]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer























          • this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
            – Zarko
            2 days ago








          • 2




            @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
            – marmot
            2 days ago










          • good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
            – Zarko
            2 days ago










          • @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
            – Joep Awinita
            2 days ago



















          6














          this is not really answer (for it i was to late for one minute), just an illustration to @marmot how i imagine rho "tail" to circle ...



          enter image description here



          documentclass[tikz, margin=3mm]{standalone}
          usetikzlibrary{chains, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 4mm and 6mm,
          start chain = going below left,
          box/.style = {minimum width=5ex, inner xsep=0pt,
          on chain, join=by latex-}
          ]
          def n {11}
          def radius {3.5cm}
          def margin {8} % margin in angles, depends on the radius
          % the cycle
          foreach s [count=i from 0,
          count=j from 1] in {456, 1562, 792, 1872, 2152,
          25, 441, 615, 2993, 2329, 2031}
          {
          node (sj) at (-i*360/n:radius) {$s$};
          draw[latex-] (i*360/n + margin:radius)
          arc (i*360/n +margin:j*360/n -margin:radius);
          }
          node (d1) [box, below left=of s8] {$2$};
          node[box] {$12$};
          node[box] {$152$};
          node[box] {$1223$};
          node[box] {$1031$};
          node[box] {$2916$};
          node[box] {$751$};
          node[box] {$1149$};
          draw[red,-latex] (d1) -- (s8);
          end{tikzpicture}

          end{document}


          as all can see, difference between codes can be neglected ... slope of line can be adjusted with node distance. for example, i would rather start at node on circle with "441" and make slope more stepped:



          node distance = 5mm and -2.5ex,


          and star with tail:



          node (d1) [box, below left=of s7] {$2$};


          for red arrows i was not sure, if it s desired (so it is red)



          enter image description here






          share|improve this answer























          • +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
            – marmot
            2 days ago










          • @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
            – Zarko
            2 days ago








          • 1




            The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
            – marmot
            2 days ago












          • oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
            – Zarko
            2 days ago






          • 1




            @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
            – marmot
            2 days ago













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          9














          Your circle and straight lines can be coded much shorter, which also allows you to change the positions of the various nodes very easily.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain},
          every join/.style=-latex]]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here



          Just for fun: the straight line of a latex rho has an angle of approximately 76 degrees.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}
          node[scale=15]{$rho$};
          draw[white,thick,double=blue] (-0.5,0.6) -- (-1.05,-1.6)
          node[midway,left=5mm,scale=3,blue]{pgfmathparse{atan2(0.6+1.6,1.05-0.5)}%
          $pgfmathprintnumber{pgfmathresult}^circ$};
          end{tikzpicture}
          end{document}


          enter image description here



          This raises the question if one can make the chain such that it has this angle. The answer is yes.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain,
          ,xshift=-{cot(76)*1.5cm}},
          every join/.style=-latex]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer























          • this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
            – Zarko
            2 days ago








          • 2




            @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
            – marmot
            2 days ago










          • good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
            – Zarko
            2 days ago










          • @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
            – Joep Awinita
            2 days ago
















          9














          Your circle and straight lines can be coded much shorter, which also allows you to change the positions of the various nodes very easily.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain},
          every join/.style=-latex]]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here



          Just for fun: the straight line of a latex rho has an angle of approximately 76 degrees.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}
          node[scale=15]{$rho$};
          draw[white,thick,double=blue] (-0.5,0.6) -- (-1.05,-1.6)
          node[midway,left=5mm,scale=3,blue]{pgfmathparse{atan2(0.6+1.6,1.05-0.5)}%
          $pgfmathprintnumber{pgfmathresult}^circ$};
          end{tikzpicture}
          end{document}


          enter image description here



          This raises the question if one can make the chain such that it has this angle. The answer is yes.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain,
          ,xshift=-{cot(76)*1.5cm}},
          every join/.style=-latex]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer























          • this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
            – Zarko
            2 days ago








          • 2




            @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
            – marmot
            2 days ago










          • good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
            – Zarko
            2 days ago










          • @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
            – Joep Awinita
            2 days ago














          9












          9








          9






          Your circle and straight lines can be coded much shorter, which also allows you to change the positions of the various nodes very easily.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain},
          every join/.style=-latex]]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here



          Just for fun: the straight line of a latex rho has an angle of approximately 76 degrees.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}
          node[scale=15]{$rho$};
          draw[white,thick,double=blue] (-0.5,0.6) -- (-1.05,-1.6)
          node[midway,left=5mm,scale=3,blue]{pgfmathparse{atan2(0.6+1.6,1.05-0.5)}%
          $pgfmathprintnumber{pgfmathresult}^circ$};
          end{tikzpicture}
          end{document}


          enter image description here



          This raises the question if one can make the chain such that it has this angle. The answer is yes.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain,
          ,xshift=-{cot(76)*1.5cm}},
          every join/.style=-latex]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer














          Your circle and straight lines can be coded much shorter, which also allows you to change the positions of the various nodes very easily.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain},
          every join/.style=-latex]]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here



          Just for fun: the straight line of a latex rho has an angle of approximately 76 degrees.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}
          node[scale=15]{$rho$};
          draw[white,thick,double=blue] (-0.5,0.6) -- (-1.05,-1.6)
          node[midway,left=5mm,scale=3,blue]{pgfmathparse{atan2(0.6+1.6,1.05-0.5)}%
          $pgfmathprintnumber{pgfmathresult}^circ$};
          end{tikzpicture}
          end{document}


          enter image description here



          This raises the question if one can make the chain such that it has this angle. The answer is yes.



          documentclass[tikz,border=3.14mm]{standalone}
          usetikzlibrary{chains,positioning}
          begin{document}

          begin{tikzpicture}[node distance=1.5cm]
          node[circle,minimum width=7cm] (circ) {};
          foreach X [count=Y] in {456,1562,792,1872,2152,25,441,615,2993,2329,2031}
          {node (cnY) at ({-(Y-2.5)*360/11}:3.5) {$X$}; }
          foreach Y [remember=Y as LastY (initially 11)]in {1,...,11}
          {draw[-latex,shorten >=4pt,shorten <=4pt] (cnLastY) to[bend left=10] (cnY);}
          begin{scope}[start chain = going below,every node/.append style={on chain,
          ,xshift=-{cot(76)*1.5cm}},
          every join/.style=-latex]
          node[below=of cn8] (n0) {2};
          draw[-latex] (cn8) -- (n0);
          node[join] (n1) {$12$};
          node[join] (n2) {$152$};
          node[join] (n3) {$1223$};
          node[join] (n4) {$1031$};
          node[join] (n5) {$2916$};
          node[join] (n6) {$751$};
          node[join] (n7) {$1149$};
          end{scope}
          end{tikzpicture}
          end{document}


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          marmotmarmot

          90.3k4104195




          90.3k4104195












          • this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
            – Zarko
            2 days ago








          • 2




            @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
            – marmot
            2 days ago










          • good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
            – Zarko
            2 days ago










          • @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
            – Joep Awinita
            2 days ago


















          • this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
            – Zarko
            2 days ago








          • 2




            @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
            – marmot
            2 days ago










          • good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
            – Zarko
            2 days ago










          • @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
            – Joep Awinita
            2 days ago
















          this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
          – Zarko
          2 days ago






          this remains me to p not to rho :-) (you beat me for some minutes :-(, it seem that i hibernate)
          – Zarko
          2 days ago






          2




          2




          @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
          – marmot
          2 days ago




          @Zarko Really? If it was a p, the line should extend further up, I think, and for a q it should be on the other side. So, since this is neither p nor q it must be a rho. ;-)
          – marmot
          2 days ago












          good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
          – Zarko
          2 days ago




          good conclusion, indeed +1. well i imagine that `rho is different. i will show my answer (don't laughing to much).
          – Zarko
          2 days ago












          @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
          – Joep Awinita
          2 days ago




          @marmot, a solution of excellence! However, it assumes the cycle will have more or less 11 nodes, which doesn't adapt too well for when the cycle is more or less say 4 nodes. See for example this new question which asks for a more general solution.
          – Joep Awinita
          2 days ago











          6














          this is not really answer (for it i was to late for one minute), just an illustration to @marmot how i imagine rho "tail" to circle ...



          enter image description here



          documentclass[tikz, margin=3mm]{standalone}
          usetikzlibrary{chains, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 4mm and 6mm,
          start chain = going below left,
          box/.style = {minimum width=5ex, inner xsep=0pt,
          on chain, join=by latex-}
          ]
          def n {11}
          def radius {3.5cm}
          def margin {8} % margin in angles, depends on the radius
          % the cycle
          foreach s [count=i from 0,
          count=j from 1] in {456, 1562, 792, 1872, 2152,
          25, 441, 615, 2993, 2329, 2031}
          {
          node (sj) at (-i*360/n:radius) {$s$};
          draw[latex-] (i*360/n + margin:radius)
          arc (i*360/n +margin:j*360/n -margin:radius);
          }
          node (d1) [box, below left=of s8] {$2$};
          node[box] {$12$};
          node[box] {$152$};
          node[box] {$1223$};
          node[box] {$1031$};
          node[box] {$2916$};
          node[box] {$751$};
          node[box] {$1149$};
          draw[red,-latex] (d1) -- (s8);
          end{tikzpicture}

          end{document}


          as all can see, difference between codes can be neglected ... slope of line can be adjusted with node distance. for example, i would rather start at node on circle with "441" and make slope more stepped:



          node distance = 5mm and -2.5ex,


          and star with tail:



          node (d1) [box, below left=of s7] {$2$};


          for red arrows i was not sure, if it s desired (so it is red)



          enter image description here






          share|improve this answer























          • +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
            – marmot
            2 days ago










          • @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
            – Zarko
            2 days ago








          • 1




            The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
            – marmot
            2 days ago












          • oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
            – Zarko
            2 days ago






          • 1




            @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
            – marmot
            2 days ago


















          6














          this is not really answer (for it i was to late for one minute), just an illustration to @marmot how i imagine rho "tail" to circle ...



          enter image description here



          documentclass[tikz, margin=3mm]{standalone}
          usetikzlibrary{chains, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 4mm and 6mm,
          start chain = going below left,
          box/.style = {minimum width=5ex, inner xsep=0pt,
          on chain, join=by latex-}
          ]
          def n {11}
          def radius {3.5cm}
          def margin {8} % margin in angles, depends on the radius
          % the cycle
          foreach s [count=i from 0,
          count=j from 1] in {456, 1562, 792, 1872, 2152,
          25, 441, 615, 2993, 2329, 2031}
          {
          node (sj) at (-i*360/n:radius) {$s$};
          draw[latex-] (i*360/n + margin:radius)
          arc (i*360/n +margin:j*360/n -margin:radius);
          }
          node (d1) [box, below left=of s8] {$2$};
          node[box] {$12$};
          node[box] {$152$};
          node[box] {$1223$};
          node[box] {$1031$};
          node[box] {$2916$};
          node[box] {$751$};
          node[box] {$1149$};
          draw[red,-latex] (d1) -- (s8);
          end{tikzpicture}

          end{document}


          as all can see, difference between codes can be neglected ... slope of line can be adjusted with node distance. for example, i would rather start at node on circle with "441" and make slope more stepped:



          node distance = 5mm and -2.5ex,


          and star with tail:



          node (d1) [box, below left=of s7] {$2$};


          for red arrows i was not sure, if it s desired (so it is red)



          enter image description here






          share|improve this answer























          • +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
            – marmot
            2 days ago










          • @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
            – Zarko
            2 days ago








          • 1




            The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
            – marmot
            2 days ago












          • oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
            – Zarko
            2 days ago






          • 1




            @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
            – marmot
            2 days ago
















          6












          6








          6






          this is not really answer (for it i was to late for one minute), just an illustration to @marmot how i imagine rho "tail" to circle ...



          enter image description here



          documentclass[tikz, margin=3mm]{standalone}
          usetikzlibrary{chains, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 4mm and 6mm,
          start chain = going below left,
          box/.style = {minimum width=5ex, inner xsep=0pt,
          on chain, join=by latex-}
          ]
          def n {11}
          def radius {3.5cm}
          def margin {8} % margin in angles, depends on the radius
          % the cycle
          foreach s [count=i from 0,
          count=j from 1] in {456, 1562, 792, 1872, 2152,
          25, 441, 615, 2993, 2329, 2031}
          {
          node (sj) at (-i*360/n:radius) {$s$};
          draw[latex-] (i*360/n + margin:radius)
          arc (i*360/n +margin:j*360/n -margin:radius);
          }
          node (d1) [box, below left=of s8] {$2$};
          node[box] {$12$};
          node[box] {$152$};
          node[box] {$1223$};
          node[box] {$1031$};
          node[box] {$2916$};
          node[box] {$751$};
          node[box] {$1149$};
          draw[red,-latex] (d1) -- (s8);
          end{tikzpicture}

          end{document}


          as all can see, difference between codes can be neglected ... slope of line can be adjusted with node distance. for example, i would rather start at node on circle with "441" and make slope more stepped:



          node distance = 5mm and -2.5ex,


          and star with tail:



          node (d1) [box, below left=of s7] {$2$};


          for red arrows i was not sure, if it s desired (so it is red)



          enter image description here






          share|improve this answer














          this is not really answer (for it i was to late for one minute), just an illustration to @marmot how i imagine rho "tail" to circle ...



          enter image description here



          documentclass[tikz, margin=3mm]{standalone}
          usetikzlibrary{chains, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 4mm and 6mm,
          start chain = going below left,
          box/.style = {minimum width=5ex, inner xsep=0pt,
          on chain, join=by latex-}
          ]
          def n {11}
          def radius {3.5cm}
          def margin {8} % margin in angles, depends on the radius
          % the cycle
          foreach s [count=i from 0,
          count=j from 1] in {456, 1562, 792, 1872, 2152,
          25, 441, 615, 2993, 2329, 2031}
          {
          node (sj) at (-i*360/n:radius) {$s$};
          draw[latex-] (i*360/n + margin:radius)
          arc (i*360/n +margin:j*360/n -margin:radius);
          }
          node (d1) [box, below left=of s8] {$2$};
          node[box] {$12$};
          node[box] {$152$};
          node[box] {$1223$};
          node[box] {$1031$};
          node[box] {$2916$};
          node[box] {$751$};
          node[box] {$1149$};
          draw[red,-latex] (d1) -- (s8);
          end{tikzpicture}

          end{document}


          as all can see, difference between codes can be neglected ... slope of line can be adjusted with node distance. for example, i would rather start at node on circle with "441" and make slope more stepped:



          node distance = 5mm and -2.5ex,


          and star with tail:



          node (d1) [box, below left=of s7] {$2$};


          for red arrows i was not sure, if it s desired (so it is red)



          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          ZarkoZarko

          122k865158




          122k865158












          • +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
            – marmot
            2 days ago










          • @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
            – Zarko
            2 days ago








          • 1




            The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
            – marmot
            2 days ago












          • oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
            – Zarko
            2 days ago






          • 1




            @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
            – marmot
            2 days ago




















          • +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
            – marmot
            2 days ago










          • @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
            – Zarko
            2 days ago








          • 1




            The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
            – marmot
            2 days ago












          • oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
            – Zarko
            2 days ago






          • 1




            @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
            – marmot
            2 days ago


















          +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
          – marmot
          2 days ago




          +1 but then my LaTeX compiler has a bug. When I compile documentclass{article} begin{document} $rho$ end{document} I get a symbol where the line on its left is close to vertical (yet it is not precisely vertical).
          – marmot
          2 days ago












          @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
          – Zarko
          2 days ago






          @marmot, you are right, i try replicate rho how i usual wrote (in old times) on blackboard. well, my handwriting is not splendid anyway ;-(. thank you! at least i now lear how to write rho :-)
          – Zarko
          2 days ago






          1




          1




          The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
          – marmot
          2 days ago






          The "correct" angle is 76 degrees, see my answer. ;-) (I hope that you practice on the blackboard until you get precisely 76 degrees. ;-)
          – marmot
          2 days ago














          oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
          – Zarko
          2 days ago




          oh, you beat me again, @marmot. i didn't check that meanwhile you edit your answer. sorry ...
          – Zarko
          2 days ago




          1




          1




          @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
          – marmot
          2 days ago






          @JoepAwinita I guess you only need to make sure that the ratio of the node distances is the arctan(76) or arccot(76) depending on how you take the ratio.
          – marmot
          2 days ago




















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