Me pregunto si puedo crear puntos como los de la imagen con tikz:
Código
\begin{tikzpicture}
\tikzstyle{level 1}=[sibling distance=60mm]
\tikzstyle{level 2}=[sibling distance=33mm]
\tikzstyle{level 3}=[sibling distance=10mm]
\tikzstyle{level 4}=[sibling distance=5mm]
\node {n}
child {node {$\frac{n}{3}$}
child {node {$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}
child {node {$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}}
child {node {$\frac{n}{3}$}
child {node {$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}
child {node {$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}}
child {node {$\frac{n}{3}$}
child{node{$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}
child{node{$\frac{n}{9}$}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}
child{node {$\frac{n}{27}$}}}
};
Y quiero comenzar las líneas de puntos en los últimos nodos secundarios y entre los nodos secundarios.
Respuesta1
Aquí hay una solución usando Tikz. El código se basó en gran medida eneste ejemplo de un árbol de recursividad, pero también muy modificado para adaptarse a su imagen de ejemplo. No estoy seguro de si querías que cn/nfueran fracciones reales, pero eso se puede solucionar fácilmente a partir del siguiente código.
Producción
Código
\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usepackage{tikz-qtree}
\usetikzlibrary{trees,calc,arrows.meta,positioning,decorations.pathreplacing,bending}
\tikzset{
edge from parent/.style={draw, thick, blue!70!black},
no edge from this parent/.style={
every child/.append style={
edge from parent/.style={draw=none}}},
level 3/.style={yshift=5cm},
level 4/.style={level distance=5mm}
}
\begin{document}
\begin{tikzpicture}[
level/.style={sibling distance=40mm/#1},
text=blue!70!black,
>=latex,
font=\sffamily
]
\node (z){cn}
child {node (a) {cn/2}
child {node (b) {cn/4}
child {node (b1) {$\vdots$}[no edge from this parent]
child {node (b11) {c}}
}
child {node (b2) {$\vdots$}[no edge from this parent]
child {node (b12) {c}}
}
}
child {node (g) {cn/4}
child {node (g1) {$\vdots$}[no edge from this parent]
child {node (g11) {c}}
}
child {node (g2) {$\vdots$}[no edge from this parent]
child {node (g12) {c}}
}
}
}
child {node (d) {cn/2}
child {node (e) {cn/4}
child {node (e1) {$\vdots$}[no edge from this parent]
child {node (e11) {c}}
}
child {node (e2) {$\vdots$}[no edge from this parent]
child {node (e12) {c}}
}
}
child {node (f) {cn/4}
child {node (f1) {$\vdots$}[no edge from this parent]
child {node (f11) {c}}
}
child {node (f2) {$\vdots$}[no edge from this parent]
child {node (f12) {c}
}
}
}
};
\node[left=5 of z] (ln1) {cn}[no edge from this parent]
child {node (ln2) {cn}[no edge from this parent]
child {node (ln3) {cn}[no edge from this parent]
child {node (ln4) {}[no edge from this parent]
child {node (ln5) {cn}}}}};
\path (b12.north east) -- (g11.north west) node [midway] {$\cdots$};
\path (e12.north east) -- (f11.north west) node [midway] {$\cdots$};
\coordinate (cd1) at ($(f12)+(1,0)$);
\coordinate (nb1) at ($(g12)!.5!(e11)$);
\draw[blue!70!black,thick,<->,]
(cd1) -- (cd1|-z.east) node [near start, fill=white] {log(n)};
\draw[blue!70!black,dashed,thick,->]
($(z.west)+(-1em,0)$) -- (ln1);
\draw[blue!70!black,dashed,thick,->]
($(a.west)+(-1em,0)$) -- (ln2.east);
\draw[blue!70!black,dashed,thick,->]
($(b.west)+(-1em,0)$) -- (ln3);
\draw[blue!70!black,dashed,thick,->]
($(b11.west)+(-1em,0)$) -- (ln5);
\draw[blue!70!black,thick,decorate,decoration={brace,amplitude=10pt,mirror},->,-{latex[flex=1pt]}] (b11.south west) -- (f12.south east);
\end{tikzpicture}
\end{document}
Respuesta2
Aquí hay una solución usando forest. Tenga en cuenta que esto toma el 'me gusta' de la imagen de destino.extremadamenteen serio. Esto significa que la especificación real del árbol principal es muy, muy compacta:
[c, name=node 1
[, name=node 2 [, name=node 3 [][]][[][]]][[[][]][[][]]]
]
Obviamente, si desea algo menos genérico, puede especificar el contenido de los nodos manualmente.
Tenga en cuenta también que realmente no hay ninguna razón para que especifique las etiquetas de los nodos 4-13 de la forma en que lo hago. En particular, la especificación de los nodos numerados del 10 al 13 en realidad se debe a que quería experimentar un poco con los recorridos por nodos.
\documentclass[tikz,border=5pt]{standalone}
\usepackage{forest,textcomp}
\usetikzlibrary{positioning,arrows.meta,decorations.pathreplacing}
\begin{document}
\begin{forest}
for tree={
font=\sffamily,
before typesetting nodes={
if={level>0}{
if n children=0{
append/.wrap pgfmath arg={
[#1, font=\sffamily, edge=dotted]
}{content("!r")},
content={},
}{
content/.wrap 2 pgfmath args={#2n\textfractionsolidus#1}{int(level()*2)}{content("!r")}
}
}{
delay n=1{
content/.wrap value={##1n},
node walk={
every step/.style={alias=node 4}, F,
every step/.style={alias=node 10}, N,
every step/.style={alias=node 11}, N,
every step/.style={alias=node 9},
node walk={
rL,
node walk={
every step/.style={alias=node 12}, P,
every step/.style={alias=node 13}, P}
}
}
}
}
},
}
[c, name=node 1
[, name=node 2 [, name=node 3 [][]][[][]]][[[][]][[][]]]
]
\begin{scope}[font=\sffamily, >=Triangle]
\node (node 5) [left=25pt of node 4] {cn};
\foreach \i/\j in {6/3,7/2,8/1} \node (node \i) at (node 5 |- node \j) {cn};
\foreach \i/\j in {1/8,2/7,3/6,4/5} \draw [densely dashed, ->] (node \i.west) -- (node \j.east);
\path [draw, decorate, decoration={brace, amplitude=10pt}] (node 9.south east) -- (node 4.south west);
\path [draw, <->] (node 9.south east) +(15pt,0) coordinate (c) -- (c |- node 1.north) node [pos=.4, fill=white] {log(n)};
\end{scope}
\draw [dotted] (node 10.north east) -- (node 11.north west) (node 13.north east) -- (node 12.north west);
\end{forest}
\end{document}
