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luku12.tex
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luku12.tex
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\chapter{Graph search}
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\chapter{Graph traversal}
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This chapter introduces two fundamental
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This chapter discusses two fundamental
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graph algorithms:
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depth-first search and breadth-first search.
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Both algorithms are given a starting
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@ -15,7 +15,7 @@ in which they visit the nodes.
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\index{depth-first search}
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\key{Depth-first search} (DFS)
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is a straightforward graph search technique.
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is a straightforward graph traversal technique.
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The algorithm begins at a starting node,
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and proceeds to all other nodes that are
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reachable from the starting node using
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@ -24,14 +24,14 @@ the edges in the graph.
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Depth-first search always follows a single
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path in the graph as long as it finds
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new nodes.
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After this, it returns back to previous
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After this, it returns to previous
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nodes and begins to explore other parts of the graph.
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The algorithm keeps track of visited nodes,
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so that it processes each node only once.
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\subsubsection*{Example}
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Let's consider how depth-first search processes
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Let us consider how depth-first search processes
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the following graph:
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\begin{center}
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\begin{tikzpicture}
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@ -92,9 +92,9 @@ After this, nodes 3 and 5 will be visited:
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\end{center}
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The neighbors of node 5 are 2 and 3,
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but the search has already visited both of them,
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so it's time to return back.
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so it is time to return to previous nodes.
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Also the neighbors of nodes 3 and 2
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have been visited, so we'll next proceed
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have been visited, so we next move
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from node 1 to node 4:
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\begin{center}
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\begin{tikzpicture}
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@ -128,7 +128,7 @@ implemented using recursion.
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The following function \texttt{dfs} begins
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a depth-first search at a given node.
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The function assumes that the graph is
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stored as adjacency lists in array
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stored as adjacency lists in an array
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\begin{lstlisting}
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vector<int> v[N];
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\end{lstlisting}
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@ -175,7 +175,7 @@ have been visited.
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\subsubsection*{Example}
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Let's consider how the algorithm processes
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Let us consider how the algorithm processes
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the following graph:
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\begin{center}
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@ -196,7 +196,7 @@ the following graph:
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\path[draw,thick,-] (5) -- (6);
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\end{tikzpicture}
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\end{center}
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Assume again that the search begins at node 1.
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Suppose again that the search begins at node 1.
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First, we process all nodes that can be reached
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from node 1 using a single edge:
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\begin{center}
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@ -224,7 +224,7 @@ from node 1 using a single edge:
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\path[draw=red,thick,->,line width=2pt] (1) -- (4);
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\end{tikzpicture}
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\end{center}
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After this, we procees to nodes 3 and 5:
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After this, we proceed to nodes 3 and 5:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle,fill=lightgray] (1) at (1,5) {$1$};
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@ -301,10 +301,10 @@ and $m$ is the number of edges.
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\subsubsection*{Implementation}
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Breadth-first search is more difficult
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to implement than depth-first search
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to implement than depth-first search,
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because the algorithm visits nodes
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in different parts in the graph.
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A typical implementation is to maintain
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in different parts of the graph.
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A typical implementation is based on
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a queue of nodes to be processed.
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At each step, the next node in the queue
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will be processed.
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@ -326,9 +326,9 @@ In addition, the code uses arrays
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\begin{lstlisting}
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int z[N], e[N];
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\end{lstlisting}
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so that array \texttt{z} indicates
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which nodes the search already has visited
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and array \texttt{e} will contain the
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so that the array \texttt{z} indicates
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which nodes the search has already visited
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and the array \texttt{e} will contain the
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minimum distance to all nodes in the graph.
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The search can be implemented as follows:
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\begin{lstlisting}
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@ -347,12 +347,12 @@ while (!q.empty()) {
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\section{Applications}
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Using the graph search algorithms,
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Using the graph traversal algorithms,
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we can check many properties of the graph.
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Usually, either depth-first search or
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bredth-first search can be used,
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but in practice, depth-first search
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is a better choice because it is
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is a better choice, because it is
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easier to implement.
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In the following applications we will
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assume that the graph is undirected.
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@ -403,18 +403,18 @@ the following nodes:
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\end{tikzpicture}
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\end{center}
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Since the search didn't visit all the nodes,
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Since the search did not visit all the nodes,
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we can conclude that the graph is not connected.
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In a similar way, we can also find all components
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in a graph by iterating trough the nodes and always
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starting a new depth-first search if the node
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doesn't belong to a component.
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In a similar way, we can also find all connected components
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of a graph by iterating trough the nodes and always
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starting a new depth-first search if the current node
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does not belong to any component yet.
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\subsubsection{Finding cycles}
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\index{cycle}
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A graph contains a cycle if during a graph search,
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A graph contains a cycle if during a graph traversal,
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we find a node whose neighbor (other than the
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previous node in the current path) has already been
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visited.
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@ -442,7 +442,7 @@ For example, the graph
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\end{center}
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contains a cycle because when we move from
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node 2 to node 5 it turns out
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that the neighbor node 3 has already been visited.
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that the neighbor 3 has already been visited.
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Thus, the graph contains a cycle that goes through node 3,
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for example, $3 \rightarrow 2 \rightarrow 5 \rightarrow 3$.
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If a component contains $c$ nodes and no cycle,
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it must contain exactly $c-1$ edges.
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If there are $c$ or more edges, the component
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always contains a cycle.
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surely contains a cycle.
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\subsubsection{Bipartiteness check}
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@ -460,9 +460,9 @@ always contains a cycle.
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A graph is bipartite if its nodes can be colored
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using two colors so that there are no adjacent
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nodes with same color.
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nodes with the same color.
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It is suprisingly easy to check if a graph
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is bipartite using graph search algorithms.
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is bipartite using graph traversal algorithms.
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The idea is to color the starting node blue,
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all its neighbors red, all their neighbors blue, and so on.
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\end{tikzpicture}
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\end{center}
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is not bipartite because a search from node 1
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produces the following situation:
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proceeds as follows:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle,fill=red!40] (2) at (5,5) {$2$};
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is red, while they are adjacent nodes in the graph.
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Thus, the graph is not bipartite.
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This algorithm always works because when there
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This algorithm always works, because when there
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are only two colors available,
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the color of the starting node in a component
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determines the colors of all other nodes in the component.
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It doesn't make any difference whether the
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It does not make any difference whether the
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starting node is red or blue.
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Note that in the general case,
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