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extra-1.tex
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\exercise
Given the following tree:
%
\begin{figure}[H]
\centering
\tikzstyle{level 1}=[level distance=1.2cm, sibling distance=1.5cm]
\tikzstyle{level 2}=[level distance=1.2cm, sibling distance=1.25cm]
\tikzstyle{level 3}=[level distance=1.2cm, sibling distance=1cm]
\begin{tikzpicture}[grow=down, sloped]
\node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {1}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {2}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {3}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {6}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {7}
edge from parent[-]
}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {4}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {5}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {8}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {9}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {11}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {12}
edge from parent[-]
}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt] {10}
edge from parent[-]
}
edge from parent[-]
};
\end{tikzpicture}
\end{figure}
%
\begin{enumerate}
\item compute its LOUDS representation;
\item compute the children of node 5;
\item can we jump directly to its fourth child?
\end{enumerate}
\solution
\begin{enumerate}
\item First, we have to place a dummy root node to the tree and, then, compute
the degrees of each note, as in picture:
%
\begin{figure}[H]
\centering
\tikzstyle{level 1}=[level distance=1.2cm, sibling distance=2cm]
\tikzstyle{level 2}=[level distance=1.2cm, sibling distance=1.5cm]
\tikzstyle{level 3}=[level distance=1.2cm, sibling distance=1cm]
\begin{tikzpicture}[grow=down, sloped]
\node[draw,fill=black, circle, inner sep=2pt, minimum size=8pt, label={\scriptsize 1}] {}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize \quad 4}] {1}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {2}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 2}] {3}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {6}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {7}
edge from parent[-]
}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {4}
edge from parent[-]
} child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 3}] {5}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {8}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize \quad 2}] {9}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {11}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {12}
edge from parent[-]
}
edge from parent[-]
}
child {
node[draw,fill=white, circle, inner sep=2pt, minimum size=17pt, label={\scriptsize 0}] {10}
edge from parent[-]
}
edge from parent[-]
}
};
\end{tikzpicture}
\end{figure}
%
We now construct its LOUDS representation by concatenating the degrees of each
node (from top to bottom, from left to right), expressed in inverse unary:
%
\begin{align*}
B = \overbracket[0pt][1pt]{10}^{root}\ \overbracket[0pt][1pt]{11110}^{1}\
\overbracket[0pt][1pt]{0}^{2}\ \overbracket[0pt][1pt]{110}^{3}\
\overbracket[0pt][1pt]{0}^{4}\ \overbracket[0pt][1pt]{1110}^{5}\
\overbracket[0pt][1pt]{0}^{6}\ \overbracket[0pt][1pt]{0}^{7}\
\overbracket[0pt][1pt]{0}^{8}\ \overbracket[0pt][1pt]{110}^{9}\
\overbracket[0pt][1pt]{0}^{10}\ \overbracket[0pt][1pt]{0}^{11}\
\overbracket[0pt][1pt]{0}^{12}\
\end{align*}
%
In this array, the $i$-th node is represented by the $i$-th 1 in the sequence,
while its children correspond to the 1s following the $i$-th 0 (so, every node
is represented \emph{twice}).
\item We cand find the children of the $i$-th node (in this case, $i = 5$)
using a $Select_0$ data structure in the following way:
%
\begin{enumerate}
\item we compute the index of the first bit after the fifth 0, with
$$y = Select_0(B, i) + 1 = Select_0(B, 5) + 1 = 13;$$
\item we compute the $id$ of the child at position $y$ with $$Rank_1(B, y) =
Select_0(B, i) - i = y - i = 8;$$
\item we search for the next sibling by scanning the next bit in the array
($y = y + 1$) and repeating the procedure from step (b) until we find that
$B_y = 0$, then we have finished.
\end{enumerate}
\item No, we have to scan the children bit-by-bit or check if the node has at
least 4 children by computing its degree with
$$degree(i) = Select_0(B, i + 1) - Select_0(B, i) - 1.$$
In this case, $degree(5) = 3 < 4$, so we can not jump to its fourth child.
\end{enumerate}