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chapter16.tex
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@ -1,7 +1,6 @@
\chapter{Directed graphs} \chapter{Directed graphs}
In this chapter, we focus on two classes of directed graphs: In this chapter, we focus on two classes of directed graphs:
\begin{itemize} \begin{itemize}
\item \key{Acyclic graphs}: \item \key{Acyclic graphs}:
There are no cycles in the graph, There are no cycles in the graph,
@ -43,7 +42,7 @@ For example, for the graph
\path[draw,thick,->,>=latex] (3) -- (6); \path[draw,thick,->,>=latex] (3) -- (6);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
a possible topological sort is one topological sort is
$[4,1,5,2,3,6]$: $[4,1,5,2,3,6]$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
@ -206,7 +205,7 @@ and there can be several topological sorts for a graph.
Let us now consider a graph for which we Let us now consider a graph for which we
cannot construct a topological sort, cannot construct a topological sort,
because there is a cycle in the graph: because the graph contains a cycle:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
@ -247,7 +246,7 @@ The search proceeds as follows:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The search reaches node 2 whose state is 1, The search reaches node 2 whose state is 1,
which means the graph contains a cycle. which means that the graph contains a cycle.
In this example, there is a cycle In this example, there is a cycle
$2 \rightarrow 3 \rightarrow 5 \rightarrow 2$. $2 \rightarrow 3 \rightarrow 5 \rightarrow 2$.
@ -296,15 +295,15 @@ There are a total of three such paths:
\item $1 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6$ \item $1 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6$
\end{itemize} \end{itemize}
Let $p(x)$ denote the number of paths from Let $\texttt{paths}(x)$ denote the number of paths from
node 1 to node $x$. node 1 to node $x$.
As a base case, $p(1)=1$. As a base case, $\texttt{paths}(1)=1$.
Then, to calculate other values of $p(x)$, Then, to calculate other values of $\texttt{paths}(x)$,
we may use the recursion we may use the recursion
\[p(x) = p(a_1)+p(a_2)+\cdots+p(a_k)\] \[\texttt{paths}(x) = \texttt{paths}(a_1)+\texttt{paths}(a_2)+\cdots+\texttt{paths}(a_k)\]
where $a_1,a_2,\ldots,a_k$ are the nodes from which there where $a_1,a_2,\ldots,a_k$ are the nodes from which there
is an edge to $x$. is an edge to $x$.
Since the graph is acyclic, the values of $p(x)$ Since the graph is acyclic, the values of $\texttt{paths}(x)$
can be calculated in the order of a topological sort. can be calculated in the order of a topological sort.
A topological sort for the above graph is as follows: A topological sort for the above graph is as follows:
\begin{center} \begin{center}
@ -352,11 +351,11 @@ Hence, the numbers of paths are as follows:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
For example, to calculate the value of $p(3)$, For example, to calculate the value of $\texttt{paths}(3)$,
we may use the formula $p(2)+p(5)$, we can use the formula $\texttt{paths}(2)+\texttt{paths}(5)$,
because there are edges from nodes 2 and 5 because there are edges from nodes 2 and 5
to node 3. to node 3.
Since $p(2)=2$ and $p(5)=1$, we conclude that $p(3)=3$. Since $\texttt{paths}(2)=2$ and $\texttt{paths}(5)=1$, we conclude that $\texttt{paths}(3)=3$.
\subsubsection{Extending Dijkstra's algorithm} \subsubsection{Extending Dijkstra's algorithm}
@ -496,7 +495,7 @@ one cycle and some paths that lead to it.
Successor graphs are sometimes called Successor graphs are sometimes called
\key{functional graphs}. \key{functional graphs}.
The reason for this is that any successor graph The reason for this is that any successor graph
corresponds to a function $f$ that defines corresponds to a function that defines
the edges of the graph. the edges of the graph.
The parameter for the function is a node of the graph, The parameter for the function is a node of the graph,
and the function gives the successor of that node. and the function gives the successor of that node.
@ -506,7 +505,7 @@ For example, the function
\begin{tabular}{r|rrrrrrrrr} \begin{tabular}{r|rrrrrrrrr}
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline \hline
$f(x)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\ $\texttt{succ}(x)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\
\end{tabular} \end{tabular}
\end{center} \end{center}
defines the following graph: defines the following graph:
@ -535,10 +534,10 @@ defines the following graph:
\end{center} \end{center}
Since each node of a successor graph has a Since each node of a successor graph has a
unique successor, we can also define a function $f(x,k)$ unique successor, we can also define a function $\texttt{succ}(x,k)$
that gives the node that we will reach if that gives the node that we will reach if
we begin at node $x$ and walk $k$ steps forward. we begin at node $x$ and walk $k$ steps forward.
For example, in the above graph $f(4,6)=2$, For example, in the above graph $\texttt{succ}(4,6)=2$,
because we will reach node 2 by walking 6 steps from node 4: because we will reach node 2 by walking 6 steps from node 4:
\begin{center} \begin{center}
@ -560,21 +559,21 @@ because we will reach node 2 by walking 6 steps from node 4:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
A straightforward way to calculate a value of $f(x,k)$ A straightforward way to calculate a value of $\texttt{succ}(x,k)$
is to start at node $x$ and walk $k$ steps forward, which takes $O(k)$ time. is to start at node $x$ and walk $k$ steps forward, which takes $O(k)$ time.
However, using preprocessing, any value of $f(x,k)$ However, using preprocessing, any value of $\texttt{succ}(x,k)$
can be calculated in only $O(\log k)$ time. can be calculated in only $O(\log k)$ time.
The idea is to precalculate all values $f(x,k)$ where The idea is to precalculate all values of $\texttt{succ}(x,k)$ where
$k$ is a power of two and at most $u$, where $u$ is $k$ is a power of two and at most $u$, where $u$ is
the maximum number of steps we will ever walk. the maximum number of steps we will ever walk.
This can be efficiently done, because This can be efficiently done, because
we can use the following recursion: we can use the following recursion:
\begin{equation*} \begin{equation*}
f(x,k) = \begin{cases} \texttt{succ}(x,k) = \begin{cases}
f(x) & k = 1\\ \texttt{succ}(x) & k = 1\\
f(f(x,k/2),k/2) & k > 1\\ \texttt{succ}(\texttt{succ}(x,k/2),k/2) & k > 1\\
\end{cases} \end{cases}
\end{equation*} \end{equation*}
@ -586,25 +585,25 @@ In the above graph, the first values are as follows:
\begin{tabular}{r|rrrrrrrrr} \begin{tabular}{r|rrrrrrrrr}
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline \hline
$f(x,1)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\ $\texttt{succ}(x,1)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\
$f(x,2)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\ $\texttt{succ}(x,2)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$f(x,4)$ & 3 & 2 & 7 & 2 & 5 & 5 & 1 & 2 & 3 \\ $\texttt{succ}(x,4)$ & 3 & 2 & 7 & 2 & 5 & 5 & 1 & 2 & 3 \\
$f(x,8)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\ $\texttt{succ}(x,8)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$\cdots$ \\ $\cdots$ \\
\end{tabular} \end{tabular}
\end{center} \end{center}
After this, any value of $f(x,k)$ can be calculated After this, any value of $\texttt{succ}(x,k)$ can be calculated
by presenting the number of steps $k$ as a sum of powers of two. by presenting the number of steps $k$ as a sum of powers of two.
For example, if we want to calculate the value of $f(x,11)$, For example, if we want to calculate the value of $\texttt{succ}(x,11)$,
we first form the representation $11=8+2+1$. we first form the representation $11=8+2+1$.
Using that, Using that,
\[f(x,11)=f(f(f(x,8),2),1).\] \[\texttt{succ}(x,11)=\texttt{succ}(\texttt{succ}(\texttt{succ}(x,8),2),1).\]
For example, in the previous graph For example, in the previous graph
\[f(4,11)=f(f(f(4,8),2),1)=5.\] \[\texttt{succ}(4,11)=\texttt{succ}(\texttt{succ}(\texttt{succ}(4,8),2),1)=5.\]
Such a representation always consists of Such a representation always consists of
$O(\log k)$ parts, so calculating a value of $f(x,k)$ $O(\log k)$ parts, so calculating a value of $\texttt{succ}(x,k)$
takes $O(\log k)$ time. takes $O(\log k)$ time.
\section{Cycle detection} \section{Cycle detection}
@ -642,7 +641,7 @@ we begin our walk at node 1,
the first node that belongs to the cycle is node 4, and the cycle consists the first node that belongs to the cycle is node 4, and the cycle consists
of three nodes (4, 5 and 6). of three nodes (4, 5 and 6).
An easy way to detect the cycle is to walk in the A simple way to detect the cycle is to walk in the
graph and keep track of graph and keep track of
all nodes that have been visited. Once a node is visited all nodes that have been visited. Once a node is visited
for the second time, we can conclude for the second time, we can conclude
@ -672,13 +671,12 @@ one step forward and the pointer $b$
walks two steps forward. walks two steps forward.
The process continues until The process continues until
the pointers meet each other: the pointers meet each other:
\begin{lstlisting} \begin{lstlisting}
a = f(x); a = succ(x);
b = f(f(x)); b = succ(succ(x));
while (a != b) { while (a != b) {
a = f(a); a = succ(a);
b = f(f(b)); b = succ(succ(b));
} }
\end{lstlisting} \end{lstlisting}
@ -688,26 +686,23 @@ so the length of the cycle divides $k$.
Thus, the first node that belongs to the cycle Thus, the first node that belongs to the cycle
can be found by moving the pointer $a$ to node $x$ can be found by moving the pointer $a$ to node $x$
and advancing the pointers and advancing the pointers
step by step until they meet again: step by step until they meet again.
\begin{lstlisting} \begin{lstlisting}
a = x; a = x;
while (a != b) { while (a != b) {
a = f(a); a = succ(a);
b = f(b); b = succ(b);
} }
first = a;
\end{lstlisting} \end{lstlisting}
Now $a$ and $b$ point to the first node in the cycle After this, the length of the cycle
that can be reached from node $x$.
Finally, the length $c$ of the cycle
can be calculated as follows: can be calculated as follows:
\begin{lstlisting} \begin{lstlisting}
b = f(a); b = succ(a);
c = 1; length = 1;
while (a != b) { while (a != b) {
b = f(b); b = succ(b);
c++; length++;
} }
\end{lstlisting} \end{lstlisting}