Add reference etc.

chapter18.tex

@@ -555,69 +555,6 @@ common ancestor of given two nodes.
 For example, in the following tree,
 the lowest common ancestor of nodes 5 and 8
 is node 2:
-\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) {$6$};
-\node[draw, circle] (8) at (-1,-3) {$8$};
-\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,-] (7) -- (8);
-\end{tikzpicture}
-\end{center}
-
-Next we will discuss two efficient techniques for
-finding the lowest common ancestor of two nodes.
-
-\subsubsection{Method 1}
-
-One way to solve the problem is to use the fact
-that we can efficiently find the $k$th
-ancestor of any node in the tree.
-Thus, we can first make sure that
-both nodes are at the same level in the tree,
-and then find the smallest value of $k$
-such that the $k$th ancestor of both nodes is the same.
-
-As an example, let us find the lowest common
-ancestor of nodes $5$ and $8$:
-
-\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,fill=lightgray] (6) at (-3,-1) {$5$};
-\node[draw, circle] (7) at (-1,-1) {$6$};
-\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
-\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,-] (7) -- (8);
-\end{tikzpicture}
-\end{center}
-
-Node $5$ is at level $3$, while node $8$ is at level $4$.
-Thus, we first move one step upwards from node $8$ to node $6$.
-After this, it turns out that the parent of both nodes $5$
-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{tikzpicture}[scale=0.9]
 \node[draw, circle] (1) at (0,3) {$1$};
@@ -637,15 +574,88 @@ The following picture shows how we move in the tree:
 \path[draw,thick,-] (7) -- (8);
 
 \path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
+\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=40] (3);
+\end{tikzpicture}
+\end{center}
+
+Next we will discuss two efficient techniques for
+finding the lowest common ancestor of two nodes.
+
+\subsubsection{Method 1}
+
+One way to solve the problem is to use the fact
+that we can efficiently find the $k$th
+ancestor of any node in the tree.
+Using this, we can divide the problem of
+finding the lowest common ancestor into two parts.
+
+We use two pointers that initially point to the
+two nodes for which we should find the
+lowest common ancestor.
+First, we move one of the pointers upwards
+so that both nodes are at the same level in the tree.
+
+In the example case, we move from node 8 to node 6,
+after which both nodes are at the same level:
+
+\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,fill=lightgray] (6) at (-3,-1) {$5$};
+\node[draw, circle,fill=lightgray] (7) at (-1,-1) {$6$};
+\node[draw, circle] (8) at (-1,-3) {$8$};
+\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,-] (7) -- (8);
+
 \path[draw=red,thick,->,line width=2pt] (8) edge [bend right] (7);
+\end{tikzpicture}
+\end{center}
+
+After this, we determine the minimum number of steps
+needed to move both pointers upwards so that
+they will point to the same node.
+This node is the lowest common ancestor of the nodes.
+
+In the example case, it suffices to move both pointers
+one step upwards to node 2,
+which is the lowest common ancestor:
+
+\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,fill=lightgray] (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) {$6$};
+\node[draw, circle] (8) at (-1,-3) {$8$};
+\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,-] (7) -- (8);
+
+\path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
 \path[draw=red,thick,->,line width=2pt] (7) edge [bend right] (3);
 \end{tikzpicture}
 \end{center}
 
-Using this method, we can find the lowest common ancestor
+Since both parts of the algorithm can be performed in
-of any two nodes in $O(\log n)$ time after an $O(n \log n)$ time
+$O(\log n)$ time using precomputed information,
-preprocessing, because both steps can be
+we can find the lowest common ancestor of any two
-performed in $O(\log n)$ time.
+nodes in $O(\log n)$ time using this technique.
 
 \subsubsection{Method 2}
 
@@ -689,13 +699,14 @@ using a depth-first search:
 \end{tikzpicture}
 \end{center}
 
-However, we use a bit different variant of
+However, we use a bit different tree
-the tree traversal array where
+traversal array than before:
 we add each node to the array \emph{always}
-when the depth-first search visits the node,
+when the depth-first search walks through the node,
-and not only at the first visit.
+and not only at the first visit\footnote{A similar technique is sometimes called the
+\key{Euler tour technique} \cite{tar84}.}.
 Hence, a node that has $k$ children appears $k+1$ times
-in the array, and there are a total of $2n-1$
+in the array and there are a total of $2n-1$
 nodes in the array.
 
 We store two values in the array: