This solution uses numerous post-anchoring adjustments to position nodes in their desired positions on the canvas. This is done using the regardless at key (see p380 of TikZ manual). The result is that rather than nodes being placed at locations determined by the graph drawing algorithm, the nodes are placed at the nominated coordinates.
In an instance like this, where the location of the nodes is known, and they have a symmetrical layout, my preference would be to use either a matrix of nodes, or the features of the positioning library to place nodes appropriately, without the need to invoke any graph drawing algorithm. Nevertheless, the question asks for a solution using graph drawing syntax, so maybe this will be useful.
To make it easier to follow the code and the graph it generates, the nodes are identified with a name (a through m), together with the labels given by the OP. This is done with a command \lb. The background grid is from How can I show coordinates by grid in TikZ automatically?
This is the result:
This is the code:
%! TEX program = lualatex
\documentclass[tikz,border=3pt]{standalone}
\usetikzlibrary{
graphs,
graphdrawing,
}
\usegdlibrary{
layered}
\pagestyle{empty}
% https://tex.stackexchange.com/a/39698
\makeatletter
\def\grd@save@target#1{%
\def\grd@target{#1}}
\def\grd@save@start#1{%
\def\grd@start{#1}}
\tikzset{
grid with coordinates/.style={
to path={%
\pgfextra{%
\edef\grd@@target{(\tikztotarget)}%
\tikz@scan@one@point\grd@save@target\grd@@target\relax
\edef\grd@@start{(\tikztostart)}%
\tikz@scan@one@point\grd@save@start\grd@@start\relax
\draw[minor help lines] (\tikztostart) grid (\tikztotarget);
\draw[major help lines] (\tikztostart) grid (\tikztotarget);
\grd@start
\pgfmathsetmacro{\grd@xa}{\the\pgf@x/1cm}
\pgfmathsetmacro{\grd@ya}{\the\pgf@y/1cm}
\grd@target
\pgfmathsetmacro{\grd@xb}{\the\pgf@x/1cm}
\pgfmathsetmacro{\grd@yb}{\the\pgf@y/1cm}
\pgfmathsetmacro{\grd@xc}{\grd@xa + \pgfkeysvalueof{/tikz/grid with coordinates/major step}}
\pgfmathsetmacro{\grd@yc}{\grd@ya + \pgfkeysvalueof{/tikz/grid with coordinates/major step}}
\foreach \x in {\grd@xa,\grd@xc,...,\grd@xb}
\node[anchor=north,color=blue] at ([yshift=-0.15cm]\x,\grd@ya) {\pgfmathprintnumber{\x}};
\foreach \y in {\grd@ya,\grd@yc,...,\grd@yb}
\node[anchor=east,color=blue] at ([xshift=-0.15cm]\grd@xa,\y) {\pgfmathprintnumber{\y}};
}
}
},
minor help lines/.style={
help lines,
step=\pgfkeysvalueof{/tikz/grid with coordinates/minor step},
minor line width/.initial=0.25pt,
},
major help lines/.style={
help lines,
line width=\pgfkeysvalueof{/tikz/grid with coordinates/major line width},
step=\pgfkeysvalueof{/tikz/grid with coordinates/major step},
line cap=round,
color=lightgray,
},
grid with coordinates/.cd,
minor step/.initial=1, % just use major
major step/.initial=1,
major line width/.initial=0.5pt,
}
\makeatother
\newcommand{\lb}[3]{{\begin{tabular}{c}#1\\#2\\\fbox{#3}\end{tabular}}}
\begin{document}
\begin{tikzpicture}[
font=\normalsize\bfseries
]
\draw (-14,-20) to[grid with coordinates] (8,1);
\graph[
layered layout,
sibling distance=50mm,
level distance=30mm,
]{
a/\lb{a}{$P4/m2/m/2/m$}{} -> b/\lb{b}{$P4/n2_1/m2/m$}{$HT-WO_3$} -> d/\lb{d}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(0,-6)}],
%%
d -> {f/\lb{f}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(-6,-9)}],g/\lb{g}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(6,-9)}]},
%%
f -> h/\lb{h}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(-6,-12)}] -> j/\lb{j}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(-6,-15)}],
g -> i/\lb{i}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(6,-12)}] -> l/\lb{l}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(6,-15)}] -> m/\lb{m}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(6,-18)}],
h -> k/\lb{k}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(0,-15)}],
k -> m,
%%
b -> c/\lb{c}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(-6,-6)}] -> e/\lb{e}{$P4/n2_1/m2/m$}{$HT-WO_3$}[regardless at={(-12,-9)}]
};
\end{tikzpicture}
\end{document}
matrix of nodes. I guess alsotikzcdcould be another approach which has been suggested for a different question of me. However, I should point you to my original question. Please have a look at the third picture. As you can see it can be become more complicated. – Hotschke Jun 05 '18 at 03:36