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chapter16.tex
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@ -1,7 +1,6 @@
\chapter{Directed graphs}
In this chapter, we focus on two classes of directed graphs:
\begin{itemize}
\item \key{Acyclic graphs}:
There are no cycles in the graph,
@ -43,7 +42,7 @@ For example, for the graph
\path[draw,thick,->,>=latex] (3) -- (6);
\end{tikzpicture}
\end{center}
a possible topological sort is
one topological sort is
$[4,1,5,2,3,6]$:
\begin{center}
\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
cannot construct a topological sort,
because there is a cycle in the graph:
because the graph contains a cycle:
\begin{center}
\begin{tikzpicture}[scale=0.9]
@ -247,7 +246,7 @@ The search proceeds as follows:
\end{tikzpicture}
\end{center}
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
$2 \rightarrow 3 \rightarrow 5 \rightarrow 2$.
@ -296,15 +295,15 @@ There are a total of three such paths:
\item $1 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6$
\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 $x$.
As a base case, $p(1)=1$.
Then, to calculate other values of $p(x)$,
As a base case, $\texttt{paths}(1)=1$.
Then, to calculate other values of $\texttt{paths}(x)$,
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
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.
A topological sort for the above graph is as follows:
\begin{center}
@ -352,11 +351,11 @@ Hence, the numbers of paths are as follows:
\end{tikzpicture}
\end{center}
For example, to calculate the value of $p(3)$,
we may use the formula $p(2)+p(5)$,
For example, to calculate the value of $\texttt{paths}(3)$,
we can use the formula $\texttt{paths}(2)+\texttt{paths}(5)$,
because there are edges from nodes 2 and 5
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}
@ -496,7 +495,7 @@ one cycle and some paths that lead to it.
Successor graphs are sometimes called
\key{functional graphs}.
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 parameter for the function is a node of the graph,
and the function gives the successor of that node.
@ -506,7 +505,7 @@ For example, the function
\begin{tabular}{r|rrrrrrrrr}
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\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{center}
defines the following graph:
@ -535,10 +534,10 @@ defines the following graph:
\end{center}
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
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:
\begin{center}
@ -560,21 +559,21 @@ because we will reach node 2 by walking 6 steps from node 4:
\end{tikzpicture}
\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.
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.
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
the maximum number of steps we will ever walk.
This can be efficiently done, because
we can use the following recursion:
\begin{equation*}
f(x,k) = \begin{cases}
f(x) & k = 1\\
f(f(x,k/2),k/2) & k > 1\\
\texttt{succ}(x,k) = \begin{cases}
\texttt{succ}(x) & k = 1\\
\texttt{succ}(\texttt{succ}(x,k/2),k/2) & k > 1\\
\end{cases}
\end{equation*}
@ -586,25 +585,25 @@ In the above graph, the first values are as follows:
\begin{tabular}{r|rrrrrrrrr}
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
$f(x,1)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\
$f(x,2)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$f(x,4)$ & 3 & 2 & 7 & 2 & 5 & 5 & 1 & 2 & 3 \\
$f(x,8)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$\texttt{succ}(x,1)$ & 3 & 5 & 7 & 6 & 2 & 2 & 1 & 6 & 3 \\
$\texttt{succ}(x,2)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$\texttt{succ}(x,4)$ & 3 & 2 & 7 & 2 & 5 & 5 & 1 & 2 & 3 \\
$\texttt{succ}(x,8)$ & 7 & 2 & 1 & 2 & 5 & 5 & 3 & 2 & 7 \\
$\cdots$ \\
\end{tabular}
\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.
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 $11=8+2+1$.
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
\[f(4,11)=f(f(f(4,8),2),1)=5.\]
\[\texttt{succ}(4,11)=\texttt{succ}(\texttt{succ}(\texttt{succ}(4,8),2),1)=5.\]
Such a representation always consists of
$O(\log k)$ 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.
\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
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
all nodes that have been visited. Once a node is visited
for the second time, we can conclude
@ -672,13 +671,12 @@ one step forward and the pointer $b$
walks two steps forward.
The process continues until
the pointers meet each other:
\begin{lstlisting}
a = f(x);
b = f(f(x));
a = succ(x);
b = succ(succ(x));
while (a != b) {
a = f(a);
b = f(f(b));
a = succ(a);
b = succ(succ(b));
}
\end{lstlisting}
@ -688,26 +686,23 @@ so the length of the cycle divides $k$.
Thus, the first node that belongs to the cycle
can be found by moving the pointer $a$ to node $x$
and advancing the pointers
step by step until they meet again:
step by step until they meet again.
\begin{lstlisting}
a = x;
while (a != b) {
a = f(a);
b = f(b);
a = succ(a);
b = succ(b);
}
first = a;
\end{lstlisting}
Now $a$ and $b$ point to the first node in the cycle
that can be reached from node $x$.
Finally, the length $c$ of the cycle
After this, the length of the cycle
can be calculated as follows:
\begin{lstlisting}
b = f(a);
c = 1;
b = succ(a);
length = 1;
while (a != b) {
b = f(b);
c++;
b = succ(b);
length++;
}
\end{lstlisting}