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@ -11,12 +11,12 @@ While Eulerian and Hamiltonian paths look like
similar concepts at first glance, similar concepts at first glance,
the computational problems related to them the computational problems related to them
are very different. are very different.
It turns out that a simple rule based on node degrees It turns out that there is a simple rule that
determines if a graph contains an Eulerian path, determines whether a graph contains an Eulerian path,
and there is also an efficient algorithm for and there is also an efficient algorithm for
finding the path. finding a path if it exists.
On the contrary, finding a Hamiltonian path is a NP-hard On the contrary, checking the existence of a Hamiltonian path is a NP-hard
problem and thus no efficient algorithm is known for solving the problem. problem and no efficient algorithm is known for solving the problem.
\section{Eulerian path} \section{Eulerian path}
@ -67,7 +67,7 @@ has an Eulerian path from node 2 to node 5:
\end{center} \end{center}
\index{Eulerian circuit} \index{Eulerian circuit}
An \key{Eulerian circuit} An \key{Eulerian circuit}
is an Eulerian path that begins and ends is an Eulerian path that starts and ends
at the same node. at the same node.
For example, the graph For example, the graph
\begin{center} \begin{center}
@ -113,12 +113,9 @@ has an Eulerian circuit that starts and ends at node 1:
\subsubsection{Existence} \subsubsection{Existence}
It turns out that the existence of Eulerian paths and circuits The existence of Eulerian paths and circuits
depends on the degrees of the nodes in the graph. depends on the degrees of the nodes in the graph.
The degree of a node is the number of its neighbours, i.e., First, an undirected graph has an Eulerian path if all the edges
the number of nodes that are connected with a direct edge.
An undirected graph has an Eulerian path if all the edges
belong to the same connected component and belong to the same connected component and
\begin{itemize} \begin{itemize}
\item the degree of each node is even \emph{or} \item the degree of each node is even \emph{or}
@ -126,9 +123,10 @@ belong to the same connected component and
and the degree of all other nodes is even. and the degree of all other nodes is even.
\end{itemize} \end{itemize}
In the first case, each Eulerian path is also an Eulerian circuit. In the first case, each Eulerian path in the graph
is also an Eulerian circuit.
In the second case, the odd-degree nodes are the starting In the second case, the odd-degree nodes are the starting
and ending nodes of an Eulerian path, and it is not an Eulerian circuit. and ending nodes of an Eulerian path which is not an Eulerian circuit.
\begin{samepage} \begin{samepage}
For example, in the graph For example, in the graph
@ -149,20 +147,15 @@ For example, in the graph
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
\end{samepage} \end{samepage}
the degree of nodes 1, 3 and 4 is 2, nodes 1, 3 and 4 have a degree of 2,
and the degree of nodes 2 and 5 is 3. and nodes 2 and 5 have a degree of 3.
Exactly two nodes have an even degree, Exactly two nodes have an even degree,
so there is an Eulerian path between nodes 2 and 5, so there is an Eulerian path between nodes 2 and 5,
but the graph doesn't contain an Eulerian circuit. but the graph does not contain an Eulerian circuit.
In a directed graph, the situation is a bit more difficult. In a directed graph,
In this case we should focus on indegree and outdegrees we should focus on indegrees and outdegrees
of the nodes in the graph. of the nodes in the graph.
The indegree of a node is the number of edges that
end at the node,
and correspondingly, the outdegree is the number of
edges that begin at the node.
A directed graph contains an Eulerian path A directed graph contains an Eulerian path
if all the edges belong to the same strongly if all the edges belong to the same strongly
connected component and connected component and
@ -173,9 +166,9 @@ in another node, outdegree is one larger than indegree,
and all other nodes have the same indegree and outdegree. and all other nodes have the same indegree and outdegree.
\end{itemize} \end{itemize}
In the first case, In the first case, each Eulerian path in the graph
each Eulerian path is also an Eulerian circuit, is also an Eulerian circuit,
and in the second case, the graph only contains an Eulerian path and in the second case, the graph contains an Eulerian path
that begins at the node whose outdegree is larger that begins at the node whose outdegree is larger
and ends at the node whose indegree is larger. and ends at the node whose indegree is larger.
@ -229,41 +222,41 @@ from node 2 to node 5:
\index{Hierholzer's algorithm} \index{Hierholzer's algorithm}
\key{Hierholzer's algorithm} constructs an Eulerian circuit \key{Hierholzer's algorithm} is an efficient
in an undirected graph. method for constructing an Eulerian circuit
for an undirected graph.
The algorithm assumes that all edges belong to The algorithm assumes that all edges belong to
the same connected component, the same connected component,
and the degree of each node is even. and the degree of each node is even.
The algorithm can be implemented in $O(n+m)$ time.
First, the algorithm constructs a circuit that contains First, the algorithm constructs a circuit that contains
some (not necessarily all) of the edges in the graph. some (not necessarily all) of the edges in the graph.
After this, the algorithm extends the circuit After this, the algorithm extends the circuit
step by step by adding subcircuits to it. step by step by adding subcircuits to it.
This continues until all edges have been added The process continues until all edges have been added
and the Eulerian circuit is ready. to the circuit.
The algorithm extends the circuit by always choosing The algorithm extends the circuit by always choosing
a node $x$ that belongs to the circuit but has a node $x$ that belongs to the circuit but has
some edges that are not included in the circuit. some edges that are not included in the circuit.
The algorith constructs a new path from node $x$ The algorithm constructs a new path from node $x$
that only contains edges that are not in the circuit. that only contains edges that are not in the circuit.
Since the degree of each node is even, Since the degree of each node is even,
sooner or later the path will return to node $x$ the path will return to the node $x$ sooner or later,
which creates a subcircuit. which creates a subcircuit.
If the graph contains two odd-degree nodes, If the graph contains two odd-degree nodes,
Hierholzer's algorithm can also be used for Hierholzer's algorithm can also be used to
constructing an Eulerian path by adding an construct an Eulerian path by adding an
extra edge between the odd-degree nodes. extra edge between the odd-degree nodes.
After this, we can first construct an Eulerian circuit After this, we can first construct an Eulerian circuit
and then remove the extra edge, and then remove the extra edge,
which produces an Eulerian path in the original graph. which yields an Eulerian path in the original graph.
\subsubsection{Example} \subsubsection{Example}
\begin{samepage} \begin{samepage}
Let's consider the following graph: Let us consider the following graph:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (3,5) {$1$}; \node[draw, circle] (1) at (3,5) {$1$};
@ -289,9 +282,9 @@ Let's consider the following graph:
\end{samepage} \end{samepage}
\begin{samepage} \begin{samepage}
Assume that the algorithm first creates a circuit Suppose the algorithm first creates a circuit
that begins at node 1. that begins at node 1.
One possible circuit is A possible circuit is
$1 \rightarrow 2 \rightarrow 3 \rightarrow 1$: $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
@ -321,8 +314,9 @@ $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$:
\end{center} \end{center}
\end{samepage} \end{samepage}
After this, the algorithm adds After this, the algorithm adds
a subcircuit the subcircuit
$2 \rightarrow 5 \rightarrow 6 \rightarrow 2$: $2 \rightarrow 5 \rightarrow 6 \rightarrow 2$
to the circuit:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (3,5) {$1$}; \node[draw, circle] (1) at (3,5) {$1$};
@ -352,8 +346,9 @@ $2 \rightarrow 5 \rightarrow 6 \rightarrow 2$:
\path[draw=red,thick,->,line width=2pt] (3) -- node[font=\small,label={[red]east:6.}] {} (1); \path[draw=red,thick,->,line width=2pt] (3) -- node[font=\small,label={[red]east:6.}] {} (1);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
Finally, the algorithm adds a subcircuit Finally, the algorithm adds the subcircuit
$6 \rightarrow 3 \rightarrow 4 \rightarrow 7 \rightarrow 6$: $6 \rightarrow 3 \rightarrow 4 \rightarrow 7 \rightarrow 6$
to the circuit:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (3,5) {$1$}; \node[draw, circle] (1) at (3,5) {$1$};
@ -467,8 +462,8 @@ that begins and ends at node 1:
\subsubsection{Existence} \subsubsection{Existence}
No efficient way is known to check if a graph No efficient method is known for testing if a graph
contains a Hamiltonian path. contains a Hamiltonian path, but the problem is NP-hard.
Still, in some special cases we can be certain Still, in some special cases we can be certain
that the graph contains a Hamiltonian path. that the graph contains a Hamiltonian path.
@ -494,8 +489,8 @@ the graph contains a Hamiltonian path.
A common feature in these theorems and other results is A common feature in these theorems and other results is
that they guarantee that a Hamiltonian path exists that they guarantee that a Hamiltonian path exists
if the graph has \emph{a lot} of edges. if the graph has \emph{a lot} of edges.
This makes sense because the more edges the graph has, This makes sense, because the more edges the graph contains,
the more possibilities we have to construct a Hamiltonian graph. the more possibilities there is to construct a Hamiltonian graph.
\subsubsection{Construction} \subsubsection{Construction}
@ -507,15 +502,14 @@ whether it exists.
A simple way to search for a Hamiltonian path is A simple way to search for a Hamiltonian path is
to use a backtracking algorithm that goes through all to use a backtracking algorithm that goes through all
possibilities how to construct the path. possible ways to construct the path.
The time complexity of such an algorithm is at least $O(n!)$, The time complexity of such an algorithm is at least $O(n!)$,
because there are $n!$ different ways to form a path because there are $n!$ different ways to choose the order of $n$ nodes.
from $n$ nodes.
A more efficient solution is based on dynamic programming A more efficient solution is based on dynamic programming
(see Chapter 10.4). (see Chapter 10.4).
The idea is to define a function $f(s,x)$, The idea is to define a function $f(s,x)$,
where $s$ is a subset of nodes, and $x$ where $s$ is a subset of nodes and $x$
is one of the nodes in the subset. is one of the nodes in the subset.
The function indicates whether there is a Hamiltonian path The function indicates whether there is a Hamiltonian path
that visits the nodes in $s$ and ends at node $x$. that visits the nodes in $s$ and ends at node $x$.
@ -567,9 +561,9 @@ The following graph corresponds to the example case:
An Eulerian path in this graph produces a string An Eulerian path in this graph produces a string
that contains all strings of length $n$. that contains all strings of length $n$.
The string contains the characters in the starting node, The string contains the characters in the starting node
and all character in the edges. and all character in the edges.
The starting node contains $n-1$ characters The starting node has $n-1$ characters
and there are $k^n$ characters in the edges, and there are $k^n$ characters in the edges,
so the length of the string is $k^n+n-1$. so the length of the string is $k^n+n-1$.
@ -579,13 +573,13 @@ so the length of the string is $k^n+n-1$.
A \key{knight's tour} is a sequence of moves A \key{knight's tour} is a sequence of moves
of a knight on an $n \times n$ chessboard of a knight on an $n \times n$ chessboard
following the rules of chess where the knight following the rules of chess such that the knight
visits each square exactly once. visits each square exactly once.
The tour is \key{closed} if the knight finally The tour is \key{closed} if the knight finally
returns to the starting square and returns to the starting square and
otherwise the tour is \key{open}. otherwise the tour is \key{open}.
For example, here's a knight's tour on a $5 \times 5$ board: For example, here is an open knight's tour on a $5 \times 5$ board:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -623,7 +617,7 @@ whose nodes represent the squares of the board,
and two nodes are connected with an edge if a knight and two nodes are connected with an edge if a knight
can move between the squares according to the rules of chess. can move between the squares according to the rules of chess.
A natural way to solve the problem is to use backtracking. A natural way to construct a knight's tour is to use backtracking.
The search can be made more efficient by using The search can be made more efficient by using
\key{heuristics} that attempts to guide the knight so that \key{heuristics} that attempts to guide the knight so that
a complete tour will be found quickly. a complete tour will be found quickly.
@ -635,14 +629,14 @@ a complete tour will be found quickly.
\key{Warnsdorff's rule} is a simple and good heuristic \key{Warnsdorff's rule} is a simple and good heuristic
for finding a knight's tour. for finding a knight's tour.
Using the rule, it is possible to efficiently find a tour Using the rule, it is possible to efficiently construct a tour
even on a large board. even on a large board.
The idea is to always move the knight so that it ends up The idea is to always move the knight so that it ends up
in a square where the number of possible moves is as in a square where the number of possible moves is as
\emph{small} as possible. \emph{small} as possible.
For example, in the following case there are five For example, in the following situation there are five
possible squares where the knight can move: possible squares to which the knight can move:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (5,5); \draw (0,0) grid (5,5);
@ -655,9 +649,9 @@ possible squares where the knight can move:
\node at (3.5,1.5) {$d$}; \node at (3.5,1.5) {$d$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
In this case, Warnsdorff's rule moves the knight to square $a$, In this situation, Warnsdorff's rule moves the knight to square $a$,
because after this choice there is only a single possible move. because after this choice, there is only a single possible move.
The other choices would move the knight to squares where The other choices would move the knight to squares where
there are three moves available. there would be three moves available.