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luku18.tex
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@ -3,9 +3,9 @@
\index{tree query} \index{tree query}
This chapter discusses techniques for This chapter discusses techniques for
efficiently processing queries related processing queries related
to subtrees and paths of a rooted tree. to subtrees and paths of a rooted tree.
For example, possible queries are: For example, such queries are:
\begin{itemize} \begin{itemize}
\item what is the $k$th ancestor of a node? \item what is the $k$th ancestor of a node?
@ -60,7 +60,7 @@ can be efficiently calculated in $O(\log k)$ time
after preprocessing. after preprocessing.
The idea is to precalculate all values $f(x,k)$ The idea is to precalculate all values $f(x,k)$
where $k$ is a power of two. where $k$ is a power of two.
For example, the values for the tree above For example, the values for the above tree
are as follows: are as follows:
\begin{center} \begin{center}
@ -373,7 +373,7 @@ can be found as follows:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
To support the queries efficiently, To answer the queries efficiently,
it suffices to store the values of the it suffices to store the values of the
nodes in a binary indexed tree or segment tree. nodes in a binary indexed tree or segment tree.
After this, we can both update a value After this, we can both update a value
@ -394,7 +394,7 @@ the root to a node
\end{itemize} \end{itemize}
For example, in the following tree, For example, in the following tree,
the sum of values from the root no node 7 is the sum of values from the root node to node 7 is
$4+5+5=14$: $4+5+5=14$:
\begin{center} \begin{center}
@ -431,8 +431,8 @@ $4+5+5=14$:
\end{center} \end{center}
We can solve this problem in a similar way as before, We can solve this problem in a similar way as before,
but each value in the last row of the node array is the sum of values but now each value in the last row of the array is the sum of values
on a path from the root to a node. on a path from the root to the node.
For example, the following array corresponds to the above tree: For example, the following array corresponds to the above tree:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -538,8 +538,8 @@ Thus, to support both the operations,
we should be able to increase all values we should be able to increase all values
in a range and retrieve a single value. in a range and retrieve a single value.
This can be done in $O(\log n)$ time This can be done in $O(\log n)$ time
(see Chapter 9.4) using a binary indexed tree using a binary indexed tree
or segment tree. or segment tree (see Chapter 9.4).
\section{Lowest common ancestor} \section{Lowest common ancestor}
@ -551,7 +551,10 @@ whose subtree contains both the nodes.
A typical problem is to efficiently process A typical problem is to efficiently process
queries that ask to find the lowest queries that ask to find the lowest
common ancestor of given two nodes. common ancestor of given two nodes.
For example, in the tree
For example, in the following tree,
the lowest common ancestor of nodes 5 and 8
is node 2:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$}; \node[draw, circle] (1) at (0,3) {$1$};
@ -571,8 +574,6 @@ For example, in the tree
\path[draw,thick,-] (7) -- (8); \path[draw,thick,-] (7) -- (8);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
the lowest common ancestor of nodes 5 and 8 is node 2,
and the lowest common ancestor of nodes 3 and 4 is node 1.
Next we will discuss two efficient techniques for Next we will discuss two efficient techniques for
finding the lowest common ancestor of two nodes. finding the lowest common ancestor of two nodes.
@ -588,7 +589,7 @@ and then find the smallest value of $k$
such that the $k$th ancestor of both nodes is the same. such that the $k$th ancestor of both nodes is the same.
As an example, let us find the lowest common As an example, let us find the lowest common
ancestor of nodes $5$ and $8$ in the following tree: ancestor of nodes $5$ and $8$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
@ -615,6 +616,8 @@ Thus, we first move one step upwards from node $8$ to node $6$.
After this, it turns out that the parent of both nodes $5$ After this, it turns out that the parent of both nodes $5$
and $6$ is node $2$, so we have found the lowest common ancestor. and $6$ is node $2$, so we have found the lowest common ancestor.
The following picture shows how we move in the tree:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$}; \node[draw, circle] (1) at (0,3) {$1$};
@ -686,7 +689,8 @@ using a depth-first search:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
However, we add each node to the node array \emph{always} However, in this problem,
we add each node to the node array \emph{always}
when the depth-first search visits the node, when the depth-first search visits the node,
and not only at the first visit. and not only at the first visit.
Hence, a node that has $k$ children appears $k+1$ times Hence, a node that has $k$ children appears $k+1$ times
@ -832,25 +836,26 @@ whose level is 2.
Thus, the lowest common ancestor of Thus, the lowest common ancestor of
nodes 5 and 8 is node 2. nodes 5 and 8 is node 2.
Using a segment tree, we can find the lowest Thus, to find the lowest common ancestor
common ancestor in $O(\log n)$ time. of two nodes it suffices to process a range
Since the array is static, the time complexity minimum query.
$O(1)$ is also possible, but this is rarely needed. Since the array is static,
In both cases, the preprocessing takes $O(n \log n)$ time. we can process such queries in $O(1)$ time
after an $O(n \log n)$ time preprocessing.
\subsubsection{Distances of nodes} \subsubsection{Distances of nodes}
Finally, let us consider a problem of Finally, let us consider the problem of
finding the distance between finding the distance between
two nodes in the tree, which equals two nodes in the tree, which equals
the length of the path between them. the length of the path between them.
It turns out that this problem reduces to It turns out that this problem reduces to
finding the lowest common ancestor of the nodes. finding the lowest common ancestor of the nodes.
First, we choose an arbitrary node for the First, we root the tree arbitrarily.
root of the tree.
After this, the distance between nodes $a$ and $b$ After this, the distance between nodes $a$ and $b$
is $d(a)+d(b)-2 \cdot d(c)$, can be calculated using the formula
\[d(a)+d(b)-2 \cdot d(c),\]
where $c$ is the lowest common ancestor of $a$ and $b$ where $c$ is the lowest common ancestor of $a$ and $b$
and $d(s)$ denotes the distance from the root node and $d(s)$ denotes the distance from the root node
to node $s$. to node $s$.
@ -881,8 +886,8 @@ For example, in the tree
\end{center} \end{center}
the lowest common ancestor of nodes 5 and 8 is node 2. the lowest common ancestor of nodes 5 and 8 is node 2.
A path from node 5 to node 8 A path from node 5 to node 8
goes first upwards from node 5 to node 2, first ascends from node 5 to node 2
and then downwards from node 2 to node 8. and then descends from node 2 to node 8.
The distances of the nodes from the root are The distances of the nodes from the root are
$d(5)=3$, $d(8)=4$ and $d(2)=2$, $d(5)=3$, $d(8)=4$ and $d(2)=2$,
so the distance between nodes 5 and 8 is so the distance between nodes 5 and 8 is