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Antti H S Laaksonen 2017-05-28 13:22:15 +03:00
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chapter14.tex
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@ -10,22 +10,24 @@ and adding any edge to a tree creates a cycle.
Moreover, there is always a unique path between any Moreover, there is always a unique path between any
two nodes of a tree. two nodes of a tree.
For example, the following tree consists of 7 nodes and 6 edges: For example, the following tree consists of 8 nodes and 7 edges:
\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) {$1$};
\node[draw, circle] (2) at (2,3) {$2$}; \node[draw, circle] (2) at (2,3) {$4$};
\node[draw, circle] (3) at (0,1) {$4$}; \node[draw, circle] (3) at (0,1) {$2$};
\node[draw, circle] (4) at (2,1) {$5$}; \node[draw, circle] (4) at (2,1) {$3$};
\node[draw, circle] (5) at (4,1) {$6$}; \node[draw, circle] (5) at (4,1) {$7$};
\node[draw, circle] (6) at (-2,3) {$7$}; \node[draw, circle] (6) at (-2,3) {$5$};
\node[draw, circle] (7) at (-2,1) {$3$}; \node[draw, circle] (7) at (-2,1) {$6$};
\node[draw, circle] (8) at (-4,1) {$8$};
\path[draw,thick,-] (1) -- (2); \path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5); \path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6); \path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7); \path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -34,7 +36,7 @@ For example, the following tree consists of 7 nodes and 6 edges:
The \key{leaves} of a tree are the nodes The \key{leaves} of a tree are the nodes
with degree 1, i.e., with only one neighbor. with degree 1, i.e., with only one neighbor.
For example, the leaves of the above tree For example, the leaves of the above tree
are nodes 3, 5, 6 and 7. are nodes 3, 5, 7 and 8.
\index{root} \index{root}
\index{rooted tree} \index{rooted tree}
@ -44,26 +46,27 @@ is appointed the \key{root} of the tree,
and all other nodes are and all other nodes are
placed underneath the root. placed underneath the root.
For example, in the following tree, For example, in the following tree,
node 1 is the root of the tree. node 1 is the root node.
\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) {$1$};
\node[draw, circle] (2) at (2,1) {$2$}; \node[draw, circle] (4) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$4$}; \node[draw, circle] (2) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$5$}; \node[draw, circle] (3) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$6$}; \node[draw, circle] (7) at (2,-1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$3$}; \node[draw, circle] (5) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$7$}; \node[draw, circle] (6) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2); \path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5); \path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6); \path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (3) -- (7); \path[draw,thick,-] (4) -- (7);
\path[draw,thick,-] (6) -- (8);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
\index{child} \index{child}
\index{parent} \index{parent}
@ -73,23 +76,34 @@ is its upper neighbor.
Each node has exactly one parent, Each node has exactly one parent,
except for the root that does not have a parent. except for the root that does not have a parent.
For example, in the above tree, For example, in the above tree,
the children of node 4 are nodes 3 and 7, the children of node 2 are nodes 5 and 6,
and the parent is node 1. and its parent is node 1.
\index{subtree} \index{subtree}
The structure of a rooted tree is \emph{recursive}: The structure of a rooted tree is \emph{recursive}:
each node of the tree acts as the root of a \key{subtree} each node of the tree acts as the root of a \key{subtree}
that contains the node itself and all other nodes that contains the node itself and all nodes
that can be reached by traversing down the tree. that are in the subtrees of its children.
For example, in the above tree, the subtree of node 4 For example, in the above tree, the subtree of node 2
consists of nodes 4, 3 and 7. consists of nodes 2, 5, 6 and 8:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (2) at (-2,1) {$2$};
\node[draw, circle] (5) at (-3,-1) {$5$};
\node[draw, circle] (6) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (6) -- (8);
\end{tikzpicture}
\end{center}
\section{Tree traversal} \section{Tree traversal}
Depth-first search and breadth-first search General graph traversal algorithms
can be used for traversing the nodes of a tree. can be used to traverse the nodes of a tree.
The traversal of a tree is easier to implement than However, the traversal of a tree is easier to implement than
that of a general graph, because that of a general graph, because
there are no cycles in the tree and it is not there are no cycles in the tree and it is not
possible to reach a node from multiple directions. possible to reach a node from multiple directions.
@ -107,7 +121,7 @@ void dfs(int s, int e) {
} }
\end{lstlisting} \end{lstlisting}
The function parameters are the current node $s$ The function is given two parameters: the current node $s$
and the previous node $e$. and the previous node $e$.
The purpose of the parameter $e$ is to make sure The purpose of the parameter $e$ is to make sure
that the search only moves to nodes that the search only moves to nodes
@ -137,8 +151,8 @@ to a leaf.
As an example, let us calculate for each node $s$ As an example, let us calculate for each node $s$
a value $\texttt{count}[s]$: the number of nodes in its subtree. a value $\texttt{count}[s]$: the number of nodes in its subtree.
The subtree contains the node itself and The subtree contains the node itself and
all nodes in the subtrees of its children. all nodes in the subtrees of its children,
Thus, we can calculate the number of nodes so we can calculate the number of nodes
recursively using the following code: recursively using the following code:
\begin{lstlisting} \begin{lstlisting}
@ -157,18 +171,17 @@ void dfs(int s, int e) {
\index{diameter} \index{diameter}
The \key{diameter} of a tree The \key{diameter} of a tree
is the maximum length of a path is the maximum length of a path between two nodes.
between two nodes of the tree. For example, consider the following tree:
For example, 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) {$1$};
\node[draw, circle] (2) at (2,3) {$2$}; \node[draw, circle] (2) at (2,3) {$4$};
\node[draw, circle] (3) at (0,1) {$4$}; \node[draw, circle] (3) at (0,1) {$2$};
\node[draw, circle] (4) at (2,1) {$5$}; \node[draw, circle] (4) at (2,1) {$3$};
\node[draw, circle] (5) at (4,1) {$6$}; \node[draw, circle] (5) at (4,1) {$7$};
\node[draw, circle] (6) at (-2,3) {$7$}; \node[draw, circle] (6) at (-2,3) {$5$};
\node[draw, circle] (7) at (-2,1) {$3$}; \node[draw, circle] (7) at (-2,1) {$6$};
\path[draw,thick,-] (1) -- (2); \path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
@ -177,37 +190,67 @@ For example, in the tree
\path[draw,thick,-] (3) -- (7); \path[draw,thick,-] (3) -- (7);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
the diameter is 4, and it corresponds to two paths: The diameter of this tree is 4,
the path between nodes 3 and 6, which corresponds to the following path:
and the path between nodes 7 and 6.
Next we will learn two efficient algorithms
for calculating the diameter of a tree.
Both algorithms calculate the diameter in $O(n)$ time.
The first algorithm is based on dynamic programming,
and the second algorithm uses two depth-first searches
to calculate the diameter.
\subsubsection{Algorithm 1}
First, we root the tree arbitrarily.
After this, we use dynamic programming
to calculate for each node $x$
the maximum length of a path that begins at some leaf,
ascends to $x$ and then descends to another leaf.
The maximum length of such a path equals the diameter of the tree.
In the example graph, a longest path begins at node 7,
ascends to node 1, and then descends to node 6:
\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) {$1$};
\node[draw, circle] (2) at (2,1) {$2$}; \node[draw, circle] (2) at (2,3) {$4$};
\node[draw, circle] (3) at (-2,1) {$4$}; \node[draw, circle] (3) at (0,1) {$2$};
\node[draw, circle] (4) at (0,1) {$5$}; \node[draw, circle] (4) at (2,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$6$}; \node[draw, circle] (5) at (4,1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$3$}; \node[draw, circle] (6) at (-2,3) {$5$};
\node[draw, circle] (7) at (-1,-1) {$7$}; \node[draw, circle] (7) at (-2,1) {$6$};
\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] (7) -- (3);
\path[draw,thick,-,color=red,line width=2pt] (3) -- (1);
\path[draw,thick,-,color=red,line width=2pt] (1) -- (2);
\path[draw,thick,-,color=red,line width=2pt] (2) -- (5);
\end{tikzpicture}
\end{center}
Note that there may be several maximum-length paths.
In the above path, we could replace node 6 with node 5
to obtain another path with length 4.
Next we will discuss two $O(n)$ time algorithms
for calculating the diameter of a tree.
The first algorithm is based on dynamic programming,
and the second algorithm uses two depth-first searches.
\subsubsection{Algorithm 1}
A general way to approach many tree problems
is to first root the tree arbitrarily.
After this, we can try to solve the problem
separately for each subtree.
Our first algorithm for calculating the diameter
is based on this idea.
An important observation is that every path
in a rooted tree has a \emph{highest point}:
the highest node that belongs to the path.
Thus, we can calculate for each node the length
of the longest path whose highest point is the node.
One of those paths corresponds to the diameter of the tree.
For example, in the following tree,
node 1 is the highest point on the path
that corresponds to the diameter:
\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] (5) at (2,-1) {$7$};
\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) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
@ -222,19 +265,30 @@ ascends to node 1, and then descends to node 6:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The algorithm first calculates for each node $x$ We calculate for each node $x$ two values:
the maximum length of a path from $x$ to a leaf. \begin{itemize}
\item $\texttt{toLeaf}(x)$: the maximum length of a path from $x$ to any leaf
\item $\texttt{maxLength}(x)$: the maximum length of a path
whose highest point is $x$
\end{itemize}
For example, in the above tree, For example, in the above tree,
a longest path from node 1 to a leaf has length 2 $\texttt{toLeaf}(1)=2$, because there is a path
(the path can be $1 \rightarrow 4 \rightarrow 3$, $1 \rightarrow 2 \rightarrow 6$,
$1 \rightarrow 4 \rightarrow 7$ or $1 \rightarrow 2 \rightarrow 6$). and $\texttt{maxLength}(1)=4$,
After this, the algorithm calculates for each node because there is a path
$x$ the maximum length of a path where $x$ $6 \rightarrow 2 \rightarrow 1 \rightarrow 4 \rightarrow 7$.
is the highest point of the path. In this case, $\texttt{maxLength}(1)$ equals the diameter.
Such a path can be found by choosing
two children with longest paths to leaves. Dynamic programming can be used to calculate the above
For example, in the above graph, values for all nodes in $O(n)$ time.
nodes 2 and 4 yield a longest path for node 1. First, to calculate $\texttt{toLeaf}(x)$,
we go through the children of $x$,
choose a child $c$ with maximum $\texttt{toLeaf}(c)$
and add one to this value.
Then, to calculate $\texttt{maxLength}(x)$,
we choose two distinct children $a$ and $b$
such that the sum $\texttt{toLeaf}(a)+\texttt{toLeaf}(b)$
is maximum and add two to this sum.
\subsubsection{Algorithm 2} \subsubsection{Algorithm 2}
@ -245,16 +299,16 @@ and find the farthest node $b$ from $a$.
Then, we find the farthest node $c$ from $b$. Then, we find the farthest node $c$ from $b$.
The diameter of the tree is the distance between $b$ and $c$. The diameter of the tree is the distance between $b$ and $c$.
In the example graph, $a$, $b$ and $c$ could be: In the following graph, $a$, $b$ and $c$ could be:
\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) {$1$};
\node[draw, circle] (2) at (2,3) {$2$}; \node[draw, circle] (2) at (2,3) {$4$};
\node[draw, circle] (3) at (0,1) {$4$}; \node[draw, circle] (3) at (0,1) {$2$};
\node[draw, circle] (4) at (2,1) {$5$}; \node[draw, circle] (4) at (2,1) {$3$};
\node[draw, circle] (5) at (4,1) {$6$}; \node[draw, circle] (5) at (4,1) {$7$};
\node[draw, circle] (6) at (-2,3) {$7$}; \node[draw, circle] (6) at (-2,3) {$5$};
\node[draw, circle] (7) at (-2,1) {$3$}; \node[draw, circle] (7) at (-2,1) {$6$};
\path[draw,thick,-] (1) -- (2); \path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
@ -262,10 +316,10 @@ In the example graph, $a$, $b$ and $c$ could be:
\path[draw,thick,-] (3) -- (6); \path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7); \path[draw,thick,-] (3) -- (7);
\node[color=red] at (2,1.6) {$a$}; \node[color=red] at (2,1.6) {$a$};
\node[color=red] at (-1.4,3) {$b$}; \node[color=red] at (-2,1.6) {$b$};
\node[color=red] at (4,1.6) {$c$}; \node[color=red] at (4,1.6) {$c$};
\path[draw,thick,-,color=red,line width=2pt] (6) -- (3); \path[draw,thick,-,color=red,line width=2pt] (7) -- (3);
\path[draw,thick,-,color=red,line width=2pt] (3) -- (1); \path[draw,thick,-,color=red,line width=2pt] (3) -- (1);
\path[draw,thick,-,color=red,line width=2pt] (1) -- (2); \path[draw,thick,-,color=red,line width=2pt] (1) -- (2);
\path[draw,thick,-,color=red,line width=2pt] (2) -- (5); \path[draw,thick,-,color=red,line width=2pt] (2) -- (5);
@ -281,12 +335,12 @@ nodes hang from it:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (2,1) {$1$}; \node[draw, circle] (1) at (2,1) {$1$};
\node[draw, circle] (2) at (4,1) {$2$}; \node[draw, circle] (2) at (4,1) {$4$};
\node[draw, circle] (3) at (0,1) {$4$}; \node[draw, circle] (3) at (0,1) {$2$};
\node[draw, circle] (4) at (2,-1) {$5$}; \node[draw, circle] (4) at (2,-1) {$3$};
\node[draw, circle] (5) at (6,1) {$6$}; \node[draw, circle] (5) at (6,1) {$7$};
\node[draw, circle] (6) at (0,-1) {$3$}; \node[draw, circle] (6) at (0,-1) {$5$};
\node[draw, circle] (7) at (-2,1) {$7$}; \node[draw, circle] (7) at (-2,1) {$6$};
\path[draw,thick,-] (1) -- (2); \path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3); \path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4); \path[draw,thick,-] (1) -- (4);
@ -312,7 +366,7 @@ The farthest node from $a$
is node $b$, node $c$ or some other node is node $b$, node $c$ or some other node
that is at least as far from node $x$. that is at least as far from node $x$.
Thus, this node is always a valid choice for Thus, this node is always a valid choice for
a starting node of a path that corresponds to the diameter. an endpoint of a path that corresponds to the diameter.
\section{Distances between nodes} \section{Distances between nodes}