# Making Tikz automatic node placement relative to the quantity of nodes

This is a follow-up question to this one: Making Tikz automatic node placement work

Especially to the post of Mark Wibrow (second one).

His solution works with 16 nodes, but I'm looking for a solution that does the same but with 32 nodes, like this:

\foreach \bytes [count=\y from 0] in {
{null},
{a,b,c,d,e},
{abcd,abce,abde,acde,bcde},
{abcde}}


Can you help me editing the following:

1. the

at ({\x+abs(2-\y)+(mod(\y, 4)==0)/2}, -\y)


that all 32 nodes are symmetrical centered.

The solution above only works for 16 nodes!

2. that my nodes a -> ab, a -> ac, a -> ad and so on are connected.

The solution

\foreach \x in {0,...,15}{
\foreach \y in {\x,...,15}{
\ifnum\x=\y
\else
\process{\x}{\y}
\ifnum\pgfmathresult=1
\draw [<->] (\x) -- (\y);
\fi
\fi
}
}


doesn't work.

I wan't to draw graph to look like this (taken from [1], p. 8)

-
Can't anybody solve my questions? –  mrbela Nov 19 '13 at 22:24
Your first problem is easily solved as the numbers of nodes in each row are the binomial coefficients 1, 5, 10, 10 and 5, and 1 which can be derived using n!/(k!*(n-k)!). So the coordinates of the nodes are ({-((5!/\y!/(5-\y)!-1)/2+\x},-\y), or something like that. –  Mark Wibrow Nov 20 '13 at 6:49

The position problem can be solved as mentioned in the comment above by calculating the binomial coefficients, like this:

\documentclass[border=0.125cm]{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\begin{document}

\begin{tikzpicture}[x=2cm,y=2cm]

\foreach \l [count=\y from 0] in {{null},
{a,b,c,d,e},
{abcd,abce,abde,acde,bcde},
{abcde}}
\foreach \m [count=\x from 0] in \l
\node [ellipse, draw, anchor=base, minimum width=1cm, minimum height=0.75cm]
(\m) at ({-(5!/\y!/(5-\y)!)/2+\x},-\y) {\m};

\end{tikzpicture}

\end{document}


And this will do the connections.

\documentclass[border=0.125cm]{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}

\newcount\incount
\def\in#1#2{\incount=0\relax\edef\marshal{\noexpand\In#1@@;#2@@;}\marshal}

\def\In#1#2;#3#4;{%
\ifx#1@%
\ifx#3@%
\incount=\ifnum0>\incount-\fi\incount%
\let\next=\relax%
\else%
\def\next{\In#1#2;#4;}%
\fi%
\else%
\test#3#4;#1**?%
\def\next{\In#2;#3#4;}%
\fi%
\next%
}
\begin{document}

\begin{tikzpicture}[declare function={nchoosek(\n,\k)=\n!/(\k!*(\n-\k)!);}, x=2cm,y=3cm]

\foreach \R [count=\y from 0, evaluate={\s=nchoosek(5,\y);}, remember=\R as \r] in {{null},
{a,b,c,d,e},
{abcd,abce,abde,acde,bcde},
{abcde}}
\foreach \C [count=\x from 0] in \R {
\node [ellipse, draw, anchor=base, minimum width=1cm, minimum height=0.75cm]
(\C) at (-\s/2+\x,-\y+1) {\C};
\foreach \c in \r {
\in{\c}{\C}
\ifnum\incount=1
\draw (\c.south) -- (\C.north);
\fi
}
}

\end{tikzpicture}

\end{document}


-
Hey Mark! First of all: Thank you very much for your help! If I want to insert your code in my file I get the error: "! Missing \$ inserted". The line is the end of the \ifnum-code. Do you understand the error or is it in my document?! –  mrbela Nov 20 '13 at 11:02
@user2438518 there was a missing \fi and an extraneous \ifnum which I've now corrected. –  Mark Wibrow Nov 20 '13 at 11:11
Now it works, thanks! For the others: Mark defined \in and \In. That made problems with my \in (element of) in math-mode. So I renamed them.. –  mrbela Nov 20 '13 at 11:55
One last stupid question: Now (and thats the last what I want to do ;) ) I want to color each node. A subset of them in green color and the other in red.. How can I define anything like "is the actual node in the set one, then draw him green. if not, draw him red". THANKS!! –  mrbela Nov 20 '13 at 11:56
@user2438518 add \C/.try to the node options. Then you can define styles, for example abc/.style={fill=blue!20}. –  Mark Wibrow Nov 20 '13 at 12:23