Improve language
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Antti H S Laaksonen
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chapter14.tex
244
chapter14.tex
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@ -10,22 +10,24 @@ and adding any edge to a tree creates a cycle.
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Moreover, there is always a unique path between any
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two nodes of a tree.
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For example, the following tree consists of 7 nodes and 6 edges:
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For example, the following tree consists of 8 nodes and 7 edges:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,3) {$2$};
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\node[draw, circle] (3) at (0,1) {$4$};
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\node[draw, circle] (4) at (2,1) {$5$};
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\node[draw, circle] (5) at (4,1) {$6$};
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\node[draw, circle] (6) at (-2,3) {$7$};
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\node[draw, circle] (7) at (-2,1) {$3$};
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\node[draw, circle] (2) at (2,3) {$4$};
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\node[draw, circle] (3) at (0,1) {$2$};
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\node[draw, circle] (4) at (2,1) {$3$};
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\node[draw, circle] (5) at (4,1) {$7$};
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\node[draw, circle] (6) at (-2,3) {$5$};
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\node[draw, circle] (7) at (-2,1) {$6$};
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\node[draw, circle] (8) at (-4,1) {$8$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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\path[draw,thick,-] (2) -- (5);
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\path[draw,thick,-] (3) -- (6);
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\path[draw,thick,-] (3) -- (7);
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\path[draw,thick,-] (7) -- (8);
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\end{tikzpicture}
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\end{center}
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@ -34,7 +36,7 @@ For example, the following tree consists of 7 nodes and 6 edges:
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The \key{leaves} of a tree are the nodes
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with degree 1, i.e., with only one neighbor.
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For example, the leaves of the above tree
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are nodes 3, 5, 6 and 7.
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are nodes 3, 5, 7 and 8.
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\index{root}
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\index{rooted tree}
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@ -44,26 +46,27 @@ is appointed the \key{root} of the tree,
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and all other nodes are
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placed underneath the root.
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For example, in the following tree,
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node 1 is the root of the tree.
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node 1 is the root node.
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,1) {$2$};
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\node[draw, circle] (3) at (-2,1) {$4$};
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\node[draw, circle] (4) at (0,1) {$5$};
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\node[draw, circle] (5) at (2,-1) {$6$};
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\node[draw, circle] (6) at (-3,-1) {$3$};
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\node[draw, circle] (7) at (-1,-1) {$7$};
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\node[draw, circle] (4) at (2,1) {$4$};
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\node[draw, circle] (2) at (-2,1) {$2$};
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\node[draw, circle] (3) at (0,1) {$3$};
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\node[draw, circle] (7) at (2,-1) {$7$};
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\node[draw, circle] (5) at (-3,-1) {$5$};
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\node[draw, circle] (6) at (-1,-1) {$6$};
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\node[draw, circle] (8) at (-1,-3) {$8$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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\path[draw,thick,-] (2) -- (5);
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\path[draw,thick,-] (3) -- (6);
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\path[draw,thick,-] (3) -- (7);
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\path[draw,thick,-] (2) -- (6);
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\path[draw,thick,-] (4) -- (7);
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\path[draw,thick,-] (6) -- (8);
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\end{tikzpicture}
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\end{center}
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\index{child}
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\index{parent}
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@ -73,23 +76,34 @@ is its upper neighbor.
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Each node has exactly one parent,
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except for the root that does not have a parent.
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For example, in the above tree,
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the children of node 4 are nodes 3 and 7,
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and the parent is node 1.
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the children of node 2 are nodes 5 and 6,
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and its parent is node 1.
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\index{subtree}
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The structure of a rooted tree is \emph{recursive}:
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each node of the tree acts as the root of a \key{subtree}
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that contains the node itself and all other nodes
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that can be reached by traversing down the tree.
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For example, in the above tree, the subtree of node 4
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consists of nodes 4, 3 and 7.
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that contains the node itself and all nodes
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that are in the subtrees of its children.
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For example, in the above tree, the subtree of node 2
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consists of nodes 2, 5, 6 and 8:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (2) at (-2,1) {$2$};
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\node[draw, circle] (5) at (-3,-1) {$5$};
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\node[draw, circle] (6) at (-1,-1) {$6$};
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\node[draw, circle] (8) at (-1,-3) {$8$};
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\path[draw,thick,-] (2) -- (5);
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\path[draw,thick,-] (2) -- (6);
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\path[draw,thick,-] (6) -- (8);
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\end{tikzpicture}
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\end{center}
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\section{Tree traversal}
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Depth-first search and breadth-first search
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can be used for traversing the nodes of a tree.
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The traversal of a tree is easier to implement than
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General graph traversal algorithms
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can be used to traverse the nodes of a tree.
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However, the traversal of a tree is easier to implement than
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that of a general graph, because
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there are no cycles in the tree and it is not
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possible to reach a node from multiple directions.
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@ -107,7 +121,7 @@ void dfs(int s, int e) {
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}
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\end{lstlisting}
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The function parameters are the current node $s$
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The function is given two parameters: the current node $s$
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and the previous node $e$.
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The purpose of the parameter $e$ is to make sure
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that the search only moves to nodes
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@ -137,8 +151,8 @@ to a leaf.
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As an example, let us calculate for each node $s$
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a value $\texttt{count}[s]$: the number of nodes in its subtree.
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The subtree contains the node itself and
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all nodes in the subtrees of its children.
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Thus, we can calculate the number of nodes
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all nodes in the subtrees of its children,
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so we can calculate the number of nodes
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recursively using the following code:
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\begin{lstlisting}
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@ -157,18 +171,17 @@ void dfs(int s, int e) {
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\index{diameter}
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The \key{diameter} of a tree
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is the maximum length of a path
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between two nodes of the tree.
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For example, in the tree
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is the maximum length of a path between two nodes.
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For example, consider the following tree:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,3) {$2$};
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\node[draw, circle] (3) at (0,1) {$4$};
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\node[draw, circle] (4) at (2,1) {$5$};
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\node[draw, circle] (5) at (4,1) {$6$};
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\node[draw, circle] (6) at (-2,3) {$7$};
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\node[draw, circle] (7) at (-2,1) {$3$};
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\node[draw, circle] (2) at (2,3) {$4$};
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\node[draw, circle] (3) at (0,1) {$2$};
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\node[draw, circle] (4) at (2,1) {$3$};
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\node[draw, circle] (5) at (4,1) {$7$};
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\node[draw, circle] (6) at (-2,3) {$5$};
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\node[draw, circle] (7) at (-2,1) {$6$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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@ -177,37 +190,67 @@ For example, in the tree
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\path[draw,thick,-] (3) -- (7);
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\end{tikzpicture}
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\end{center}
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the diameter is 4, and it corresponds to two paths:
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the path between nodes 3 and 6,
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and the path between nodes 7 and 6.
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Next we will learn two efficient algorithms
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for calculating the diameter of a tree.
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Both algorithms calculate the diameter in $O(n)$ time.
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The first algorithm is based on dynamic programming,
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and the second algorithm uses two depth-first searches
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to calculate the diameter.
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\subsubsection{Algorithm 1}
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First, we root the tree arbitrarily.
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After this, we use dynamic programming
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to calculate for each node $x$
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the maximum length of a path that begins at some leaf,
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ascends to $x$ and then descends to another leaf.
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The maximum length of such a path equals the diameter of the tree.
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In the example graph, a longest path begins at node 7,
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ascends to node 1, and then descends to node 6:
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The diameter of this tree is 4,
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which corresponds to the following path:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,1) {$2$};
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\node[draw, circle] (3) at (-2,1) {$4$};
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\node[draw, circle] (4) at (0,1) {$5$};
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\node[draw, circle] (5) at (2,-1) {$6$};
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\node[draw, circle] (6) at (-3,-1) {$3$};
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\node[draw, circle] (7) at (-1,-1) {$7$};
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\node[draw, circle] (2) at (2,3) {$4$};
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\node[draw, circle] (3) at (0,1) {$2$};
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\node[draw, circle] (4) at (2,1) {$3$};
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\node[draw, circle] (5) at (4,1) {$7$};
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\node[draw, circle] (6) at (-2,3) {$5$};
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\node[draw, circle] (7) at (-2,1) {$6$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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\path[draw,thick,-] (2) -- (5);
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\path[draw,thick,-] (3) -- (6);
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\path[draw,thick,-] (3) -- (7);
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\path[draw,thick,-,color=red,line width=2pt] (7) -- (3);
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\path[draw,thick,-,color=red,line width=2pt] (3) -- (1);
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\path[draw,thick,-,color=red,line width=2pt] (1) -- (2);
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\path[draw,thick,-,color=red,line width=2pt] (2) -- (5);
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\end{tikzpicture}
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\end{center}
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Note that there may be several maximum-length paths.
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In the above path, we could replace node 6 with node 5
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to obtain another path with length 4.
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Next we will discuss two $O(n)$ time algorithms
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for calculating the diameter of a tree.
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The first algorithm is based on dynamic programming,
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and the second algorithm uses two depth-first searches.
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\subsubsection{Algorithm 1}
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A general way to approach many tree problems
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is to first root the tree arbitrarily.
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After this, we can try to solve the problem
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separately for each subtree.
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Our first algorithm for calculating the diameter
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is based on this idea.
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An important observation is that every path
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in a rooted tree has a \emph{highest point}:
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the highest node that belongs to the path.
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Thus, we can calculate for each node the length
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of the longest path whose highest point is the node.
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One of those paths corresponds to the diameter of the tree.
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For example, in the following tree,
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node 1 is the highest point on the path
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that corresponds to the diameter:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,1) {$4$};
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\node[draw, circle] (3) at (-2,1) {$2$};
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\node[draw, circle] (4) at (0,1) {$3$};
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\node[draw, circle] (5) at (2,-1) {$7$};
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\node[draw, circle] (6) at (-3,-1) {$5$};
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\node[draw, circle] (7) at (-1,-1) {$6$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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@ -222,19 +265,30 @@ ascends to node 1, and then descends to node 6:
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\end{tikzpicture}
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\end{center}
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The algorithm first calculates for each node $x$
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the maximum length of a path from $x$ to a leaf.
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We calculate for each node $x$ two values:
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\begin{itemize}
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\item $\texttt{toLeaf}(x)$: the maximum length of a path from $x$ to any leaf
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\item $\texttt{maxLength}(x)$: the maximum length of a path
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whose highest point is $x$
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\end{itemize}
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For example, in the above tree,
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a longest path from node 1 to a leaf has length 2
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(the path can be $1 \rightarrow 4 \rightarrow 3$,
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$1 \rightarrow 4 \rightarrow 7$ or $1 \rightarrow 2 \rightarrow 6$).
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After this, the algorithm calculates for each node
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$x$ the maximum length of a path where $x$
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is the highest point of the path.
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Such a path can be found by choosing
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two children with longest paths to leaves.
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For example, in the above graph,
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nodes 2 and 4 yield a longest path for node 1.
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$\texttt{toLeaf}(1)=2$, because there is a path
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$1 \rightarrow 2 \rightarrow 6$,
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and $\texttt{maxLength}(1)=4$,
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because there is a path
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$6 \rightarrow 2 \rightarrow 1 \rightarrow 4 \rightarrow 7$.
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In this case, $\texttt{maxLength}(1)$ equals the diameter.
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Dynamic programming can be used to calculate the above
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values for all nodes in $O(n)$ time.
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First, to calculate $\texttt{toLeaf}(x)$,
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we go through the children of $x$,
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choose a child $c$ with maximum $\texttt{toLeaf}(c)$
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and add one to this value.
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Then, to calculate $\texttt{maxLength}(x)$,
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we choose two distinct children $a$ and $b$
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such that the sum $\texttt{toLeaf}(a)+\texttt{toLeaf}(b)$
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is maximum and add two to this sum.
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\subsubsection{Algorithm 2}
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@ -245,16 +299,16 @@ and find the farthest node $b$ from $a$.
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Then, we find the farthest node $c$ from $b$.
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The diameter of the tree is the distance between $b$ and $c$.
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In the example graph, $a$, $b$ and $c$ could be:
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In the following graph, $a$, $b$ and $c$ could be:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,3) {$1$};
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\node[draw, circle] (2) at (2,3) {$2$};
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\node[draw, circle] (3) at (0,1) {$4$};
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\node[draw, circle] (4) at (2,1) {$5$};
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\node[draw, circle] (5) at (4,1) {$6$};
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\node[draw, circle] (6) at (-2,3) {$7$};
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\node[draw, circle] (7) at (-2,1) {$3$};
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\node[draw, circle] (2) at (2,3) {$4$};
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\node[draw, circle] (3) at (0,1) {$2$};
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\node[draw, circle] (4) at (2,1) {$3$};
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\node[draw, circle] (5) at (4,1) {$7$};
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\node[draw, circle] (6) at (-2,3) {$5$};
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\node[draw, circle] (7) at (-2,1) {$6$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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@ -262,10 +316,10 @@ In the example graph, $a$, $b$ and $c$ could be:
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\path[draw,thick,-] (3) -- (6);
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\path[draw,thick,-] (3) -- (7);
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\node[color=red] at (2,1.6) {$a$};
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\node[color=red] at (-1.4,3) {$b$};
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\node[color=red] at (-2,1.6) {$b$};
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\node[color=red] at (4,1.6) {$c$};
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\path[draw,thick,-,color=red,line width=2pt] (6) -- (3);
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\path[draw,thick,-,color=red,line width=2pt] (7) -- (3);
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\path[draw,thick,-,color=red,line width=2pt] (3) -- (1);
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\path[draw,thick,-,color=red,line width=2pt] (1) -- (2);
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\path[draw,thick,-,color=red,line width=2pt] (2) -- (5);
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@ -281,12 +335,12 @@ nodes hang from it:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (2,1) {$1$};
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\node[draw, circle] (2) at (4,1) {$2$};
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\node[draw, circle] (3) at (0,1) {$4$};
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\node[draw, circle] (4) at (2,-1) {$5$};
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\node[draw, circle] (5) at (6,1) {$6$};
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\node[draw, circle] (6) at (0,-1) {$3$};
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\node[draw, circle] (7) at (-2,1) {$7$};
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\node[draw, circle] (2) at (4,1) {$4$};
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\node[draw, circle] (3) at (0,1) {$2$};
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\node[draw, circle] (4) at (2,-1) {$3$};
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\node[draw, circle] (5) at (6,1) {$7$};
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\node[draw, circle] (6) at (0,-1) {$5$};
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\node[draw, circle] (7) at (-2,1) {$6$};
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\path[draw,thick,-] (1) -- (2);
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\path[draw,thick,-] (1) -- (3);
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\path[draw,thick,-] (1) -- (4);
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@ -312,7 +366,7 @@ The farthest node from $a$
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is node $b$, node $c$ or some other node
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that is at least as far from node $x$.
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Thus, this node is always a valid choice for
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a starting node of a path that corresponds to the diameter.
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an endpoint of a path that corresponds to the diameter.
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\section{Distances between nodes}
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Reference in New Issue