Improve 14.3
This commit is contained in:
Antti H S Laaksonen
parent
eb73621637
commit
38c59da9ea
209
chapter14.tex
209
chapter14.tex
|
|
@ -368,152 +368,163 @@ that is at least as far from node $x$.
|
|||
Thus, this node is always a valid choice for
|
||||
an endpoint of a path that corresponds to the diameter.
|
||||
|
||||
\section{Distances between nodes}
|
||||
\section{All longest paths}
|
||||
|
||||
A more difficult problem is to calculate
|
||||
for each node of the tree and for each direction,
|
||||
the maximum distance to a node in that direction.
|
||||
It turns out that this can be calculated in
|
||||
$O(n)$ time using dynamic programming.
|
||||
Our next problem is to calculate for every node
|
||||
in the tree the maximum length of a path
|
||||
that begins at the node.
|
||||
This can be seen as a generalization of the
|
||||
tree diameter problem, because the largest of those
|
||||
lengths equals the diameter of the tree.
|
||||
Also this problem can be solved in $O(n)$ time.
|
||||
|
||||
\begin{samepage}
|
||||
In the example graph, the distances are as follows:
|
||||
As an example, consider the following tree:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[draw, circle] (1) at (0,3) {$1$};
|
||||
\node[draw, circle] (2) at (2,3) {$2$};
|
||||
\node[draw, circle] (3) at (0,1) {$4$};
|
||||
\node[draw, circle] (4) at (2,1) {$5$};
|
||||
\node[draw, circle] (5) at (4,1) {$6$};
|
||||
\node[draw, circle] (6) at (-2,3) {$7$};
|
||||
\node[draw, circle] (7) at (-2,1) {$3$};
|
||||
\node[draw, circle] (1) at (0,0) {$1$};
|
||||
\node[draw, circle] (2) at (-1.5,-1) {$4$};
|
||||
\node[draw, circle] (3) at (2,0) {$2$};
|
||||
\node[draw, circle] (4) at (-1.5,1) {$3$};
|
||||
\node[draw, circle] (6) at (3.5,-1) {$6$};
|
||||
\node[draw, circle] (7) at (3.5,1) {$5$};
|
||||
\path[draw,thick,-] (1) -- (2);
|
||||
\path[draw,thick,-] (1) -- (3);
|
||||
\path[draw,thick,-] (1) -- (4);
|
||||
\path[draw,thick,-] (2) -- (5);
|
||||
\path[draw,thick,-] (3) -- (6);
|
||||
\path[draw,thick,-] (3) -- (7);
|
||||
\node[color=red] at (0.5,3.2) {$2$};
|
||||
\node[color=red] at (0.3,2.4) {$1$};
|
||||
\node[color=red] at (-0.2,2.4) {$2$};
|
||||
\node[color=red] at (-0.2,1.5) {$3$};
|
||||
\node[color=red] at (-0.5,1.2) {$1$};
|
||||
\node[color=red] at (-1.7,2.4) {$4$};
|
||||
\node[color=red] at (-0.5,0.8) {$1$};
|
||||
\node[color=red] at (-1.5,0.8) {$4$};
|
||||
\node[color=red] at (1.5,3.2) {$3$};
|
||||
\node[color=red] at (1.5,1.2) {$3$};
|
||||
\node[color=red] at (3.5,1.2) {$4$};
|
||||
\node[color=red] at (2.2,2.4) {$1$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{samepage}
|
||||
For example, the farthest node from node 4
|
||||
in the direction of node 1 is node 6, and the distance to that
|
||||
node is 3 using the path
|
||||
$4 \rightarrow 1 \rightarrow 2 \rightarrow 6$.
|
||||
|
||||
\begin{samepage}
|
||||
Let $\texttt{maxLength}(x)$ denote the maximum length
|
||||
of a path that begins at node $x$.
|
||||
For example, in the above tree,
|
||||
$\texttt{maxLength}(4)=3$, because there
|
||||
is a path $4 \rightarrow 1 \rightarrow 2 \rightarrow 6$.
|
||||
Here is a complete table of the values:
|
||||
\begin{center}
|
||||
\begin{tabular}{l|lllllll}
|
||||
node $x$ & 1 & 2 & 3 & 4 & 5 & 6 \\
|
||||
$\texttt{maxLength}(x)$ & 2 & 2 & 3 & 3 & 3 & 3 \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
Also in this problem, a good starting point
|
||||
is to root the tree.
|
||||
After this, all distances to leaves can
|
||||
be calculated using dynamic programming:
|
||||
for solving the problem is to root the tree arbitrarily:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[draw, circle] (1) at (0,3) {$1$};
|
||||
\node[draw, circle] (2) at (2,1) {$2$};
|
||||
\node[draw, circle] (3) at (-2,1) {$4$};
|
||||
\node[draw, circle] (4) at (0,1) {$5$};
|
||||
\node[draw, circle] (5) at (2,-1) {$6$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$3$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$7$};
|
||||
\node[draw, circle] (2) at (2,1) {$4$};
|
||||
\node[draw, circle] (3) at (-2,1) {$2$};
|
||||
\node[draw, circle] (4) at (0,1) {$3$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$5$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$6$};
|
||||
\path[draw,thick,-] (1) -- (2);
|
||||
\path[draw,thick,-] (1) -- (3);
|
||||
\path[draw,thick,-] (1) -- (4);
|
||||
\path[draw,thick,-] (2) -- (5);
|
||||
\path[draw,thick,-] (3) -- (6);
|
||||
\path[draw,thick,-] (3) -- (7);
|
||||
|
||||
\node[color=red] at (-2.5,0.7) {$1$};
|
||||
\node[color=red] at (-1.5,0.7) {$1$};
|
||||
|
||||
\node[color=red] at (2.2,0.5) {$1$};
|
||||
|
||||
\node[color=red] at (-0.5,2.8) {$2$};
|
||||
\node[color=red] at (0.2,2.5) {$1$};
|
||||
\node[color=red] at (0.5,2.8) {$2$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{samepage}
|
||||
|
||||
The remaining task is to calculate
|
||||
the distances through parents.
|
||||
This can be done by traversing the tree once again
|
||||
and keeping track of the largest distance from the parent
|
||||
of the current node to some other node in another direction.
|
||||
|
||||
For example, the distance from node 2 upwards
|
||||
is one larger than the distance from node 1
|
||||
downwards in some other direction than node 2:
|
||||
The first part of the problem is to calculate for every node $x$
|
||||
the maximum length of a path that goes through a child of $x$.
|
||||
For example, the longest path from node 1
|
||||
goes through its child 2:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[draw, circle] (1) at (0,3) {$1$};
|
||||
\node[draw, circle] (2) at (2,1) {$2$};
|
||||
\node[draw, circle] (3) at (-2,1) {$4$};
|
||||
\node[draw, circle] (4) at (0,1) {$5$};
|
||||
\node[draw, circle] (5) at (2,-1) {$6$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$3$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$7$};
|
||||
\node[draw, circle] (2) at (2,1) {$4$};
|
||||
\node[draw, circle] (3) at (-2,1) {$2$};
|
||||
\node[draw, circle] (4) at (0,1) {$3$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$5$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$6$};
|
||||
\path[draw,thick,-] (1) -- (2);
|
||||
\path[draw,thick,-] (1) -- (3);
|
||||
\path[draw,thick,-] (1) -- (4);
|
||||
\path[draw,thick,-] (2) -- (5);
|
||||
\path[draw,thick,-] (3) -- (6);
|
||||
\path[draw,thick,-] (3) -- (7);
|
||||
|
||||
\path[draw,thick,-,color=red,line width=2pt] (1) -- (2);
|
||||
\path[draw,thick,-,color=red,line width=2pt] (1) -- (3);
|
||||
\path[draw,thick,-,color=red,line width=2pt] (3) -- (7);
|
||||
|
||||
\path[draw,thick,->,color=red,line width=2pt] (1) -- (3);
|
||||
\path[draw,thick,->,color=red,line width=2pt] (3) -- (6);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
Finally, we can calculate the distances for all nodes
|
||||
and all directions:
|
||||
This part is easy to solve in $O(n)$ time, because we can use
|
||||
dynamic programming as we have done previously.
|
||||
|
||||
Then, the second part of the problem is to calculate
|
||||
for every node $x$ the maximum length of a path
|
||||
through its parent $p$.
|
||||
For example, the longest path
|
||||
from node 3 goes through its parent 1:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[draw, circle] (1) at (0,3) {$1$};
|
||||
\node[draw, circle] (2) at (2,1) {$2$};
|
||||
\node[draw, circle] (3) at (-2,1) {$4$};
|
||||
\node[draw, circle] (4) at (0,1) {$5$};
|
||||
\node[draw, circle] (5) at (2,-1) {$6$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$3$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$7$};
|
||||
\node[draw, circle] (2) at (2,1) {$4$};
|
||||
\node[draw, circle] (3) at (-2,1) {$2$};
|
||||
\node[draw, circle] (4) at (0,1) {$3$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$5$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$6$};
|
||||
\path[draw,thick,-] (1) -- (2);
|
||||
\path[draw,thick,-] (1) -- (3);
|
||||
\path[draw,thick,-] (1) -- (4);
|
||||
\path[draw,thick,-] (2) -- (5);
|
||||
\path[draw,thick,-] (3) -- (6);
|
||||
\path[draw,thick,-] (3) -- (7);
|
||||
|
||||
\node[color=red] at (-2.5,0.7) {$1$};
|
||||
\node[color=red] at (-1.5,0.7) {$1$};
|
||||
|
||||
\node[color=red] at (2.2,0.5) {$1$};
|
||||
|
||||
\node[color=red] at (-0.5,2.8) {$2$};
|
||||
\node[color=red] at (0.2,2.5) {$1$};
|
||||
\node[color=red] at (0.5,2.8) {$2$};
|
||||
|
||||
\node[color=red] at (-3,-0.4) {$4$};
|
||||
\node[color=red] at (-1,-0.4) {$4$};
|
||||
\node[color=red] at (-2,1.6) {$3$};
|
||||
\node[color=red] at (2,1.6) {$3$};
|
||||
|
||||
\node[color=red] at (2.2,-0.4) {$4$};
|
||||
\node[color=red] at (0.2,1.6) {$3$};
|
||||
\path[draw,thick,->,color=red,line width=2pt] (4) -- (1);
|
||||
\path[draw,thick,->,color=red,line width=2pt] (1) -- (3);
|
||||
\path[draw,thick,->,color=red,line width=2pt] (3) -- (6);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
At first glance, it seems that we should choose
|
||||
the longest path from $p$.
|
||||
However, this \emph{does not} always work,
|
||||
because the longest path from $p$
|
||||
may go through $x$.
|
||||
Here is an example of this situation:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.9]
|
||||
\node[draw, circle] (1) at (0,3) {$1$};
|
||||
\node[draw, circle] (2) at (2,1) {$4$};
|
||||
\node[draw, circle] (3) at (-2,1) {$2$};
|
||||
\node[draw, circle] (4) at (0,1) {$3$};
|
||||
\node[draw, circle] (6) at (-3,-1) {$5$};
|
||||
\node[draw, circle] (7) at (-1,-1) {$6$};
|
||||
\path[draw,thick,-] (1) -- (2);
|
||||
\path[draw,thick,-] (1) -- (3);
|
||||
\path[draw,thick,-] (1) -- (4);
|
||||
\path[draw,thick,-] (3) -- (6);
|
||||
\path[draw,thick,-] (3) -- (7);
|
||||
|
||||
\path[draw,thick,->,color=red,line width=2pt] (3) -- (1);
|
||||
\path[draw,thick,->,color=red,line width=2pt] (1) -- (2);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Still, we can solve the second part in
|
||||
$O(n)$ time by storing \emph{two} maximum lengths
|
||||
for each node $x$:
|
||||
\begin{itemize}
|
||||
\item $\texttt{maxLength}_1(x)$:
|
||||
the maximum length of a path from $x$
|
||||
\item $\texttt{maxLength}_2(x)$
|
||||
the maximum length of a path from $x$
|
||||
in another direction than the first path
|
||||
\end{itemize}
|
||||
For example, in the above graph,
|
||||
$\texttt{maxLength}_1(1)=2$
|
||||
using the path $1 \rightarrow 2 \rightarrow 5$,
|
||||
and $\texttt{maxLength}_2(1)=1$
|
||||
using the path $1 \rightarrow 3$.
|
||||
|
||||
Finally, if the path that corresponds to
|
||||
$\texttt{maxLength}_1(p)$ goes through $x$,
|
||||
we conclude that the maximum length is
|
||||
$\texttt{maxLength}_2(p)+1$,
|
||||
and otherwise the maximum length is
|
||||
$\texttt{maxLength}_1(p)+1$.
|
||||
|
||||
|
||||
\section{Binary trees}
|
||||
|
||||
\index{binary tree}
|
||||
|
|
|
|||
Loading…
Reference in New Issue