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Path queries

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Antti H S Laaksonen 2017-02-08 22:45:17 +02:00
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luku18.tex
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@ -383,18 +383,19 @@ and calculate the sum of values in $O(\log n)$ time.
Using a node array, we can also efficiently
calculate sums of values on
paths between the root node and any other
paths from the root node to any other
node in the tree.
Let us next consider a problem where our task
is to support the following queries:
\begin{itemize}
\item change the value of a node
\item calculate the sum of values on a path between
the root node and a node
\item calculate the sum of values on a path from
the root to a node
\end{itemize}
For example, in the following tree, the sum of
values from the root to node 8 is $4+5+3=12$.
For example, in the following tree,
the sum of values from the root no node 7 is
$4+5+5=14$:
\begin{center}
\begin{tikzpicture}[scale=0.9]
@ -429,21 +430,13 @@ values from the root to node 8 is $4+5+3=12$.
\end{tikzpicture}
\end{center}
To solve this problem, we can use a similar
technique as we used for subtree queries,
but the values of the nodes are stored
in a special way:
if the value of a node at position $k$
increases by $a$,
the value at position $k$ increases by $a$
and the value at position $k+c$ decreases by $a$,
where $c$ is the size of the subtree.
\begin{samepage}
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
on a path from the root to a node.
For example, the following array corresponds to the above tree:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (10,-2);
\draw (0,1) grid (9,-2);
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
@ -454,7 +447,6 @@ For example, the following array corresponds to the above tree:
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};
\node at (9.5,0.5) {--};
\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
@ -465,18 +457,16 @@ For example, the following array corresponds to the above tree:
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};
\node at (9.5,-0.5) {--};
\node at (0.5,-1.5) {$4$};
\node at (1.5,-1.5) {$5$};
\node at (2.5,-1.5) {$3$};
\node at (3.5,-1.5) {$-5$};
\node at (4.5,-1.5) {$2$};
\node at (5.5,-1.5) {$5$};
\node at (6.5,-1.5) {$-2$};
\node at (7.5,-1.5) {$-2$};
\node at (8.5,-1.5) {$-4$};
\node at (9.5,-1.5) {$-4$};
\node at (1.5,-1.5) {$9$};
\node at (2.5,-1.5) {$12$};
\node at (3.5,-1.5) {$7$};
\node at (4.5,-1.5) {$9$};
\node at (5.5,-1.5) {$14$};
\node at (6.5,-1.5) {$12$};
\node at (7.5,-1.5) {$10$};
\node at (8.5,-1.5) {$6$};
\footnotesize
\node at (0.5,1.4) {$1$};
@ -488,30 +478,18 @@ For example, the following array corresponds to the above tree:
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\node at (9.5,1.4) {$10$};
\end{tikzpicture}
\end{center}
\end{samepage}
For example, the value of node $3$ is $-5$,
because it is the next node after the subtrees
of nodes $2$ and $6$ and its own value is $3$.
So the value decreases by $5+3$ and increases by $3$.
Note that the array contains an extra index 10
that only has the opposite number of the value
of the root node.
Using this array, the sum of values in a path
from the root to node $x$ equals the sum
of values in the array from the beginning to node $x$.
For example, the sum from the root to node $8$
can be calculated as follows:
When the value of a node increases by $x$,
the sums of all nodes in its subtree increase by $x$.
For example, if the value of node 4 increases by 1,
the node array changes as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (6,1) rectangle (7,0);
\fill[color=lightgray] (0,-1) rectangle (7,-2);
\draw (0,1) grid (10,-2);
\fill[color=lightgray] (4,-1) rectangle (8,-2);
\draw (0,1) grid (9,-2);
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
@ -522,7 +500,6 @@ can be calculated as follows:
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};
\node at (9.5,0.5) {--};
\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
@ -533,18 +510,16 @@ can be calculated as follows:
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};
\node at (9.5,-0.5) {--};
\node at (0.5,-1.5) {$4$};
\node at (1.5,-1.5) {$5$};
\node at (2.5,-1.5) {$3$};
\node at (3.5,-1.5) {$-5$};
\node at (4.5,-1.5) {$2$};
\node at (5.5,-1.5) {$5$};
\node at (6.5,-1.5) {$-2$};
\node at (7.5,-1.5) {$-2$};
\node at (8.5,-1.5) {$-4$};
\node at (9.5,-1.5) {$-4$};
\node at (1.5,-1.5) {$9$};
\node at (2.5,-1.5) {$12$};
\node at (3.5,-1.5) {$7$};
\node at (4.5,-1.5) {$10$};
\node at (5.5,-1.5) {$15$};
\node at (6.5,-1.5) {$13$};
\node at (7.5,-1.5) {$11$};
\node at (8.5,-1.5) {$6$};
\footnotesize
\node at (0.5,1.4) {$1$};
@ -556,22 +531,15 @@ can be calculated as follows:
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\node at (9.5,1.4) {$10$};
\end{tikzpicture}
\end{center}
The sum is
\[4+5+3-5+2+5-2=12\]
that equals the sum $4+5+3=12$.
This method works, because the value of each node
is added to the sum when the depth-first search
visits the node for the first time, and the value
of the node is removed from the sum when the subtree of the
node has been processed.
Once again, we can store the values of the nodes
in a binary indexed tree or a segment tree,
so it is possible to both update a value and
calculate the sum of values efficiently in $O(\log n)$ time.
Thus, to support both the operations,
we should be able to increase all values
in a range and retrieve a single value.
This can be done in $O(\log n)$ time
(see Chapter 9.4) using a binary indexed tree
or segment tree.
\section{Lowest common ancestor}