823 lines
27 KiB
TeX
823 lines
27 KiB
TeX
\chapter{Shortest paths}
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\index{shortest path}
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Finding the shortest path between two nodes
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is an important graph problem that has many
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applications in practice.
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For example, a natural problem in a road network
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is to calculate the length of the shorthest route
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between two cities, given the lengths of the roads.
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In an unweighted graph, the length of a path equals
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the number of edges in the path and we can
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simply use breadth-first search for finding
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the shortest path.
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However, in this chapter we concentrate on
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weighted graphs.
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In this case we need more sophisticated algorithms
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for finding shortest paths.
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\section{Bellman–Ford algorithm}
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\index{Bellman–Ford algorithm}
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The \key{Bellman–Fordin algoritmi} finds the
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shortest path from a starting node to all
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other nodes in the graph.
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The algorithm works in all kinds of graphs,
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provided that the graph doesn't contain a
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cycle with negative length.
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If the graph contains a negative cycle,
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the algorithm can detect this.
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The algorithm keeps track of estimated distances
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from the starting node to other nodes.
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Initially, the estimated distance is 0
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to the starting node and infinite to all other nodes.
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The algorithm improves the estimates by finding
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edges that shorten the paths until it is not
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possible to improve any estimate.
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\subsubsection{Example}
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Let's consider how the Bellman–Ford algorithm
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works in the following graph:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle] (1) at (1,3) {1};
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\node[draw, circle] (2) at (4,3) {2};
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\node[draw, circle] (3) at (1,1) {3};
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\node[draw, circle] (4) at (4,1) {4};
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\node[draw, circle] (5) at (6,2) {5};
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\node[color=red] at (1,3+0.55) {$0$};
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\node[color=red] at (4,3+0.55) {$\infty$};
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\node[color=red] at (1,1-0.55) {$\infty$};
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\node[color=red] at (4,1-0.55) {$\infty$};
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\node[color=red] at (6,2-0.55) {$\infty$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
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\path[draw,thick,-] (1) -- node[font=\small,label=left:3] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-2$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=left:3] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (5);
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\path[draw,thick,-] (4) -- node[font=\small,label=below:2] {} (5);
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\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (4);
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\end{tikzpicture}
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\end{center}
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Each node in the graph is assigned an estimated distance.
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Initially, the distance is 0 to the starting node
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and infinite to all other nodes.
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The algorithm searches for edges that improve the
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estimated distances.
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First, all edges from node 1 improve the estimates:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle] (1) at (1,3) {1};
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\node[draw, circle] (2) at (4,3) {2};
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\node[draw, circle] (3) at (1,1) {3};
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\node[draw, circle] (4) at (4,1) {4};
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\node[draw, circle] (5) at (6,2) {5};
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\node[color=red] at (1,3+0.55) {$0$};
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\node[color=red] at (4,3+0.55) {$2$};
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\node[color=red] at (1,1-0.55) {$3$};
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\node[color=red] at (4,1-0.55) {$7$};
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\node[color=red] at (6,2-0.55) {$\infty$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
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\path[draw,thick,-] (1) -- node[font=\small,label=left:3] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-2$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=left:3] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (5);
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\path[draw,thick,-] (4) -- node[font=\small,label=below:2] {} (5);
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\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (4);
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\path[draw=red,thick,->,line width=2pt] (1) -- (2);
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\path[draw=red,thick,->,line width=2pt] (1) -- (3);
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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, edges
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$2 \rightarrow 5$ and $3 \rightarrow 4$
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improve the estimates:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle] (1) at (1,3) {1};
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\node[draw, circle] (2) at (4,3) {2};
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\node[draw, circle] (3) at (1,1) {3};
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\node[draw, circle] (4) at (4,1) {4};
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\node[draw, circle] (5) at (6,2) {5};
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\node[color=red] at (1,3+0.55) {$0$};
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\node[color=red] at (4,3+0.55) {$2$};
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\node[color=red] at (1,1-0.55) {$3$};
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\node[color=red] at (4,1-0.55) {$1$};
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\node[color=red] at (6,2-0.55) {$7$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
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\path[draw,thick,-] (1) -- node[font=\small,label=left:3] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-2$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=left:3] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (5);
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\path[draw,thick,-] (4) -- node[font=\small,label=below:2] {} (5);
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\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (4);
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\path[draw=red,thick,->,line width=2pt] (2) -- (5);
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\path[draw=red,thick,->,line width=2pt] (3) -- (4);
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\end{tikzpicture}
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\end{center}
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Finally, there is one more improvment:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle] (1) at (1,3) {1};
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\node[draw, circle] (2) at (4,3) {2};
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\node[draw, circle] (3) at (1,1) {3};
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\node[draw, circle] (4) at (4,1) {4};
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\node[draw, circle] (5) at (6,2) {5};
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\node[color=red] at (1,3+0.55) {$0$};
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\node[color=red] at (4,3+0.55) {$2$};
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\node[color=red] at (1,1-0.55) {$3$};
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\node[color=red] at (4,1-0.55) {$1$};
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\node[color=red] at (6,2-0.55) {$3$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
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\path[draw,thick,-] (1) -- node[font=\small,label=left:3] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-2$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=left:3] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (5);
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\path[draw,thick,-] (4) -- node[font=\small,label=below:2] {} (5);
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\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (4);
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\path[draw=red,thick,->,line width=2pt] (4) -- (5);
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\end{tikzpicture}
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\end{center}
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After this, no edge improves the estimates.
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This means that the distances are final
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and we have successfully
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calculated the shortest distance
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from the starting node to all other nodes.
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For example, the smallest distance 3
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from node 1 to node 5 corresponds to
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the following path:
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\begin{center}
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\begin{tikzpicture}
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\node[draw, circle] (1) at (1,3) {1};
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\node[draw, circle] (2) at (4,3) {2};
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\node[draw, circle] (3) at (1,1) {3};
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\node[draw, circle] (4) at (4,1) {4};
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\node[draw, circle] (5) at (6,2) {5};
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\node[color=red] at (1,3+0.55) {$0$};
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\node[color=red] at (4,3+0.55) {$2$};
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\node[color=red] at (1,1-0.55) {$3$};
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\node[color=red] at (4,1-0.55) {$1$};
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\node[color=red] at (6,2-0.55) {$3$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
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\path[draw,thick,-] (1) -- node[font=\small,label=left:3] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-2$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=left:3] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (5);
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\path[draw,thick,-] (4) -- node[font=\small,label=below:2] {} (5);
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\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (4);
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\path[draw=red,thick,->,line width=2pt] (1) -- (3);
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\path[draw=red,thick,->,line width=2pt] (3) -- (4);
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\path[draw=red,thick,->,line width=2pt] (4) -- (5);
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\end{tikzpicture}
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\end{center}
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\subsubsection{Implementation}
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The following implementation of the
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Bellman–Ford algorithm finds the shortest paths
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from a node $x$ to all other nodes in the graph.
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The code assumes that the graph is stored
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as adjacency lists in array
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\begin{lstlisting}
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vector<pair<int,int>> v[N];
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\end{lstlisting}
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so that each pair contains the target node
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and the edge weight.
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The algorithm consists of $n-1$ rounds,
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and on each round the algorithm goes through
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all nodes in the graph and tries to improve
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the estimated distances.
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The algorithm builds an array \texttt{e}
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that will contain the distance from $x$
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to all nodes in the graph.
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The initial value $10^9$ means infinity.
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\begin{lstlisting}
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for (int i = 1; i <= n; i++) e[i] = 1e9;
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e[x] = 0;
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for (int i = 1; i <= n-1; i++) {
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for (int a = 1; a <= n; a++) {
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for (auto b : v[a]) {
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e[b.first] = min(e[b.first],e[a]+b.second);
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}
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}
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}
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\end{lstlisting}
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The time complexity of the algorithm is $O(nm)$
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because it consists of $n-1$ rounds and
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iterates through all $m$ nodes during a round.
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If there are no negative cycles in the graph,
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all distances are final after $n-1$ rounds
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because each shortest path can contain at most $n-1$ edges.
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In practice, the final distances can usually
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be found much faster than in $n-1$ rounds.
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Thus, a possible way to make the algorithm more efficient
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is to stop the algorithm if we can't
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improve any distance during a round.
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\subsubsection{Negative cycle}
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\index{negative cycle}
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Using the Bellman–Ford algorithm we can also
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check if the graph contains a cycle with negative length.
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For example, the graph
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,0) {$1$};
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\node[draw, circle] (2) at (2,1) {$2$};
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\node[draw, circle] (3) at (2,-1) {$3$};
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\node[draw, circle] (4) at (4,0) {$4$};
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\path[draw,thick,-] (1) -- node[font=\small,label=above:$3$] {} (2);
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\path[draw,thick,-] (2) -- node[font=\small,label=above:$1$] {} (4);
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\path[draw,thick,-] (1) -- node[font=\small,label=below:$5$] {} (3);
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\path[draw,thick,-] (3) -- node[font=\small,label=below:$-7$] {} (4);
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\path[draw,thick,-] (2) -- node[font=\small,label=right:$2$] {} (3);
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\end{tikzpicture}
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\end{center}
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\noindent
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contains a negative cycle
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$2 \rightarrow 3 \rightarrow 4 \rightarrow 2$
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with length $-4$.
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If the graph contains a negative cycle,
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we can shorten a path that contains the cycle
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infinitely many times by repeating the cycle
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again and again.
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Thus, the concept of a shortest path
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is not meaningful here.
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A negative cycle can be detected
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using the Bellman–Ford algorithm by
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running the algorithm for $n$ rounds.
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If the last round improves any distance,
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the graph contains a negative cycle.
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Note that this algorithm searches for
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a negative cycle in the whole graph
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regardless of the starting node.
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\subsubsection{SPFA algorithm}
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\index{SPFA algorithm}
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The \key{SPFA algoritmi} (''Shortest Path Faster Algorithm'')
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is a variation for the Bellman–Ford algorithm,
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that is often more efficient than the original algorithm.
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It doesn't go through all the edges on each round,
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but instead, it chooses the edges to be examined
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in a more intelligent way.
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The algorithm maintains a queue of nodes that might
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be used for improving the distances.
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First, the algorithm adds the starting node $x$
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to the queue.
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Then, the algorithm always processes the
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first node in the queue, and when an edge
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$a \rightarrow b$ improves a distance,
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node $b$ is added to the end of the queue.
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The following implementation uses a
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\texttt{queue} structure \texttt{q}.
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In addition, array \texttt{z} indicates
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if a node is already in the queue,
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in which case the algorithm doesn't add
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the node to the queue again.
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\begin{lstlisting}
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for (int i = 1; i <= n; i++) e[i] = 1e9;
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e[x] = 0;
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q.push(x);
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while (!q.empty()) {
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int a = q.front(); q.pop();
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z[a] = 0;
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for (auto b : v[a]) {
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if (e[a]+b.second < e[b.first]) {
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e[b.first] = e[a]+b.second;
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if (!z[b]) {q.push(b); z[b] = 1;}
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}
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}
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}
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\end{lstlisting}
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The efficiency of the SPFA algorithm depends
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on the structure of the graph:
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the algorithm is usually very efficient,
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but its worst case time complexity is still
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$O(nm)$ and it is possible to create inputs
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that make the algorithm as slow as the
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standard Bellman–Ford algorithm.
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\section{Dijkstran algoritmi}
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\index{Dijkstran algoritmi@Dijkstran algoritmi}
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\key{Dijkstran algoritmi} etsii Bellman–Fordin
|
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algoritmin tavoin lyhimmät polut
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alkusolmusta kaikkiin muihin solmuihin.
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Dijkstran algoritmi on tehokkaampi kuin
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Bellman–Fordin algoritmi,
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minkä ansiosta se soveltuu suurten
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verkkojen käsittelyyn.
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Algoritmi vaatii kuitenkin,
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ettei verkossa ole negatiivisia kaaria.
|
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Dijkstran algoritmi vastaa
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Bellman–Fordin algoritmia siinä,
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että se pitää
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yllä etäisyysarvioita solmuihin
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ja parantaa niitä algoritmin aikana.
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Algoritmin tehokkuus perustuu
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siihen, että sen riittää käydä läpi
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verkon kaaret vain kerran
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hyödyntäen tietoa,
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ettei verkossa ole negatiivisia kaaria.
|
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\subsubsection{Esimerkki}
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Tarkastellaan Dijkstran algoritmin toimintaa
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seuraavassa verkossa, kun alkusolmuna
|
||
on solmu 1:
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (1,3) {3};
|
||
\node[draw, circle] (2) at (4,3) {4};
|
||
\node[draw, circle] (3) at (1,1) {2};
|
||
\node[draw, circle] (4) at (4,1) {1};
|
||
\node[draw, circle] (5) at (6,2) {5};
|
||
|
||
\node[color=red] at (1,3+0.6) {$\infty$};
|
||
\node[color=red] at (4,3+0.6) {$\infty$};
|
||
\node[color=red] at (1,1-0.6) {$\infty$};
|
||
\node[color=red] at (4,1-0.6) {$0$};
|
||
\node[color=red] at (6,2-0.6) {$\infty$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:6] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
\end{tikzpicture}
|
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\end{center}
|
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Bellman–Fordin algoritmin tavoin
|
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alkusolmun etäisyysarvio on 0
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ja kaikissa muissa solmuissa etäisyysarvio
|
||
on aluksi ääretön.
|
||
|
||
Dijkstran algoritmi
|
||
ottaa joka askeleella käsittelyyn
|
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sellaisen solmun,
|
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jota ei ole vielä käsitelty
|
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ja jonka etäisyysarvio on
|
||
mahdollisimman pieni.
|
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Alussa tällainen solmu on solmu 1,
|
||
jonka etäisyysarvio on 0.
|
||
|
||
Kun solmu tulee käsittelyyn,
|
||
algoritmi käy läpi kaikki
|
||
siitä lähtevät kaaret ja
|
||
parantaa etäisyysarvioita
|
||
niiden avulla:
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle] (1) at (1,3) {3};
|
||
\node[draw, circle] (2) at (4,3) {4};
|
||
\node[draw, circle] (3) at (1,1) {2};
|
||
\node[draw, circle, fill=lightgray] (4) at (4,1) {1};
|
||
\node[draw, circle] (5) at (6,2) {5};
|
||
|
||
\node[color=red] at (1,3+0.6) {$\infty$};
|
||
\node[color=red] at (4,3+0.6) {$9$};
|
||
\node[color=red] at (1,1-0.6) {$5$};
|
||
\node[color=red] at (4,1-0.6) {$0$};
|
||
\node[color=red] at (6,2-0.6) {$1$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:6] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
|
||
\path[draw=red,thick,->,line width=2pt] (4) -- (2);
|
||
\path[draw=red,thick,->,line width=2pt] (4) -- (3);
|
||
\path[draw=red,thick,->,line width=2pt] (4) -- (5);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
Solmun 1 käsittely paransi etäisyysarvioita
|
||
solmuihin 2, 4 ja 5,
|
||
joiden uudet etäisyydet ovat nyt 5, 9 ja 1.
|
||
|
||
Seuraavaksi käsittelyyn tulee solmu 5,
|
||
jonka etäisyys on 1:
|
||
\begin{center}
|
||
\begin{tikzpicture}
|
||
\node[draw, circle] (1) at (1,3) {3};
|
||
\node[draw, circle] (2) at (4,3) {4};
|
||
\node[draw, circle] (3) at (1,1) {2};
|
||
\node[draw, circle, fill=lightgray] (4) at (4,1) {1};
|
||
\node[draw, circle, fill=lightgray] (5) at (6,2) {5};
|
||
|
||
\node[color=red] at (1,3+0.6) {$\infty$};
|
||
\node[color=red] at (4,3+0.6) {$3$};
|
||
\node[color=red] at (1,1-0.6) {$5$};
|
||
\node[color=red] at (4,1-0.6) {$0$};
|
||
\node[color=red] at (6,2-0.6) {$1$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:6] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
|
||
\path[draw=red,thick,->,line width=2pt] (5) -- (2);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
Tämän jälkeen vuorossa on solmu 4:
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle] (1) at (1,3) {3};
|
||
\node[draw, circle, fill=lightgray] (2) at (4,3) {4};
|
||
\node[draw, circle] (3) at (1,1) {2};
|
||
\node[draw, circle, fill=lightgray] (4) at (4,1) {1};
|
||
\node[draw, circle, fill=lightgray] (5) at (6,2) {5};
|
||
|
||
\node[color=red] at (1,3+0.6) {$9$};
|
||
\node[color=red] at (4,3+0.6) {$3$};
|
||
\node[color=red] at (1,1-0.6) {$5$};
|
||
\node[color=red] at (4,1-0.6) {$0$};
|
||
\node[color=red] at (6,2-0.6) {$1$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:6] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
|
||
\path[draw=red,thick,->,line width=2pt] (2) -- (1);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
|
||
Dijkstran algoritmissa on hienoutena,
|
||
että aina kun solmu tulee käsittelyyn,
|
||
sen etäisyysarvio on siitä lähtien lopullinen.
|
||
Esimerkiksi tässä vaiheessa
|
||
etäisyydet 0, 1 ja 3 ovat lopulliset
|
||
etäisyydet solmuihin 1, 5 ja 4.
|
||
|
||
Algoritmi käsittelee vastaavasti
|
||
vielä kaksi viimeistä solmua,
|
||
minkä jälkeen algoritmin päätteeksi
|
||
etäisyydet ovat:
|
||
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle, fill=lightgray] (1) at (1,3) {3};
|
||
\node[draw, circle, fill=lightgray] (2) at (4,3) {4};
|
||
\node[draw, circle, fill=lightgray] (3) at (1,1) {2};
|
||
\node[draw, circle, fill=lightgray] (4) at (4,1) {1};
|
||
\node[draw, circle, fill=lightgray] (5) at (6,2) {5};
|
||
|
||
\node[color=red] at (1,3+0.6) {$7$};
|
||
\node[color=red] at (4,3+0.6) {$3$};
|
||
\node[color=red] at (1,1-0.6) {$5$};
|
||
\node[color=red] at (4,1-0.6) {$0$};
|
||
\node[color=red] at (6,2-0.6) {$1$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:6] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
|
||
\subsubsection{Negatiiviset kaaret}
|
||
|
||
Dijkstran algoritmin tehokkuus perustuu siihen,
|
||
että verkossa ei ole negatiivisia kaaria.
|
||
Jos verkossa on negatiivinen kaari,
|
||
algoritmi ei välttämättä toimi oikein.
|
||
Tarkastellaan esimerkkinä seuraavaa verkkoa:
|
||
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle] (1) at (0,0) {$1$};
|
||
\node[draw, circle] (2) at (2,1) {$2$};
|
||
\node[draw, circle] (3) at (2,-1) {$3$};
|
||
\node[draw, circle] (4) at (4,0) {$4$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:2] {} (2);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:3] {} (4);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=below:6] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:$-5$] {} (4);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
\noindent
|
||
Lyhin polku solmusta 1 solmuun 4 on
|
||
$1 \rightarrow 3 \rightarrow 4$,
|
||
ja sen pituus on 1.
|
||
Dijkstran algoritmi löytää
|
||
kuitenkin keveimpiä kaaria seuraten
|
||
polun $1 \rightarrow 2 \rightarrow 4$.
|
||
Algoritmi ei pysty ottamaan huomioon,
|
||
että alemmalla polulla kaaren paino $-5$
|
||
kumoaa aiemman suuren kaaren painon $6$.
|
||
|
||
\subsubsection{Toteutus}
|
||
|
||
Seuraava Dijkstran algoritmin toteutus laskee
|
||
pienimmän etäisyyden solmusta $x$ kaikkiin muihin solmuihin.
|
||
Verkko on tallennettu taulukkoon \texttt{v}
|
||
vieruslistoina, joissa on pareina kohdesolmu
|
||
ja kaaren pituus.
|
||
|
||
Dijkstran algoritmin tehokas toteutus vaatii,
|
||
että verkosta pystyy löytämään
|
||
nopeasti vielä käsittelemättömän solmun,
|
||
jonka etäisyysarvio on pienin.
|
||
Sopiva tietorakenne tähän on prioriteettijono,
|
||
jossa solmut ovat järjestyksessä etäisyys\-arvioiden mukaan.
|
||
Prioriteettijonon avulla
|
||
seuraavaksi käsiteltävän solmun saa selville logaritmisessa ajassa.
|
||
|
||
Seuraavassa toteutuksessa prioriteettijono sisältää
|
||
pareja, joiden ensimmäinen kenttä on etäisyysarvio
|
||
ja toinen kenttä on solmun tunniste:
|
||
\begin{lstlisting}
|
||
priority_queue<pair<int,int>> q;
|
||
\end{lstlisting}
|
||
Pieni hankaluus on,
|
||
että Dijkstran algoritmissa täytyy saada selville
|
||
\emph{pienimmän} etäisyysarvion solmu,
|
||
kun taas C++:n prioriteettijono antaa oletuksena
|
||
\emph{suurimman} alkion.
|
||
Helppo ratkaisu on tallentaa etäisyysarviot
|
||
\emph{negatiivisina}, jolloin C++:n prioriteettijonoa
|
||
voi käyttää suoraan.
|
||
|
||
Koodi merkitsee taulukkoon \texttt{z},
|
||
onko solmu käsitelty,
|
||
ja pitää yllä etäisyysarvioita taulukossa \texttt{e}.
|
||
Alussa alkusolmun etäisyysarvio on 0
|
||
ja jokaisen muun solmun etäisyysarviona
|
||
on ääretöntä vastaava $10^9$.
|
||
|
||
\begin{lstlisting}
|
||
for (int i = 1; i <= n; i++) e[i] = 1e9;
|
||
e[x] = 0;
|
||
q.push({0,x});
|
||
while (!q.empty()) {
|
||
int a = q.top().second; q.pop();
|
||
if (z[a]) continue;
|
||
z[a] = 1;
|
||
for (auto b : v[a]) {
|
||
if (e[a]+b.second < e[b]) {
|
||
e[b] = e[a]+b.second;
|
||
q.push({-e[b],b});
|
||
}
|
||
}
|
||
}
|
||
\end{lstlisting}
|
||
|
||
Yllä olevan toteutuksen aikavaativuus on $O(n+m \log m)$,
|
||
koska algoritmi käy läpi kaikki verkon solmut
|
||
ja lisää jokaista kaarta kohden korkeintaan
|
||
yhden etäisyysarvion prioriteettijonoon.
|
||
|
||
\section{Floyd–Warshallin algoritmi}
|
||
|
||
\index{Floyd–Warshallin algoritmi}
|
||
|
||
\key{Floyd–Warshallin algoritmi}
|
||
on toisenlainen lähestymistapa
|
||
lyhimpien polkujen etsintään.
|
||
Toisin kuin muut tämän luvun algoritmit,
|
||
se etsii yhdellä kertaa lyhimmät polut kaikkien
|
||
verkon solmujen välillä.
|
||
|
||
|
||
Algoritmi ylläpitää kaksiulotteista
|
||
taulukkoa etäisyyksistä solmujen
|
||
välillä.
|
||
Ensin taulukkoon on merkitty
|
||
etäisyydet käyttäen vain solmujen
|
||
välisiä kaaria.
|
||
Tämän jälkeen algoritmi
|
||
päivittää etäisyyksiä,
|
||
kun verkon solmut saavat yksi kerrallaan
|
||
toimia välisolmuina poluilla.
|
||
|
||
\subsubsection{Esimerkki}
|
||
|
||
Tarkastellaan Floyd–Warshallin
|
||
algoritmin toimintaa seuraavassa verkossa:
|
||
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle] (1) at (1,3) {$3$};
|
||
\node[draw, circle] (2) at (4,3) {$4$};
|
||
\node[draw, circle] (3) at (1,1) {$2$};
|
||
\node[draw, circle] (4) at (4,1) {$1$};
|
||
\node[draw, circle] (5) at (6,2) {$5$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
|
||
Algoritmi merkitsee aluksi taulukkoon
|
||
etäisyyden 0 jokaisesta solmusta itseensä
|
||
sekä etäisyyden $x$, jos solmuparin välillä
|
||
on kaari, jonka pituus on $x$.
|
||
Muiden solmuparien etäisyys on aluksi ääretön.
|
||
|
||
Tässä verkossa taulukosta tulee:
|
||
\begin{center}
|
||
\begin{tabular}{r|rrrrr}
|
||
& 1 & 2 & 3 & 4 & 5 \\
|
||
\hline
|
||
1 & 0 & 5 & $\infty$ & 9 & 1 \\
|
||
2 & 5 & 0 & 2 & $\infty$ & $\infty$ \\
|
||
3 & $\infty$ & 2 & 0 & 7 & $\infty$ \\
|
||
4 & 9 & $\infty$ & 7 & 0 & 2 \\
|
||
5 & 1 & $\infty$ & $\infty$ & 2 & 0 \\
|
||
\end{tabular}
|
||
\end{center}
|
||
\vspace{10pt}
|
||
Algoritmin toiminta muodostuu peräkkäisistä kierroksista.
|
||
Jokaisella kierroksella valitaan yksi uusi solmu,
|
||
joka saa toimia välisolmuna poluilla,
|
||
ja algoritmi parantaa taulukon
|
||
etäisyyksiä muodostaen polkuja tämän solmun avulla.
|
||
|
||
Ensimmäisellä kierroksella solmu 1 on välisolmu.
|
||
Tämän ansiosta solmujen 2 ja 4 välille muodostuu
|
||
polku, jonka pituus on 14,
|
||
koska solmu 1 yhdistää ne toisiinsa.
|
||
Vastaavasti solmut 2 ja 5 yhdistyvät polulla,
|
||
jonka pituus on 6.
|
||
|
||
\begin{center}
|
||
\begin{tabular}{r|rrrrr}
|
||
& 1 & 2 & 3 & 4 & 5 \\
|
||
\hline
|
||
1 & 0 & 5 & $\infty$ & 9 & 1 \\
|
||
2 & 5 & 0 & 2 & \textbf{14} & \textbf{6} \\
|
||
3 & $\infty$ & 2 & 0 & 7 & $\infty$ \\
|
||
4 & 9 & \textbf{14} & 7 & 0 & 2 \\
|
||
5 & 1 & \textbf{6} & $\infty$ & 2 & 0 \\
|
||
\end{tabular}
|
||
\end{center}
|
||
\vspace{10pt}
|
||
|
||
Toisella kierroksella solmu 2 saa toimia välisolmuna.
|
||
Tämä mahdollistaa uudet polut solmuparien 1 ja 3
|
||
sekä 3 ja 5 välille:
|
||
|
||
\begin{center}
|
||
\begin{tabular}{r|rrrrr}
|
||
& 1 & 2 & 3 & 4 & 5 \\
|
||
\hline
|
||
1 & 0 & 5 & \textbf{7} & 9 & 1 \\
|
||
2 & 5 & 0 & 2 & 14 & 6 \\
|
||
3 & \textbf{7} & 2 & 0 & 7 & \textbf{8} \\
|
||
4 & 9 & 14 & 7 & 0 & 2 \\
|
||
5 & 1 & 6 & \textbf{8} & 2 & 0 \\
|
||
\end{tabular}
|
||
\end{center}
|
||
\vspace{10pt}
|
||
|
||
Kolmannella kierroksella solmu 3 saa toimia välisolmuna,
|
||
jolloin syntyy uusi polku solmuparin 2 ja 4 välille:
|
||
|
||
\begin{center}
|
||
\begin{tabular}{r|rrrrr}
|
||
& 1 & 2 & 3 & 4 & 5 \\
|
||
\hline
|
||
1 & 0 & 5 & 7 & 9 & 1 \\
|
||
2 & 5 & 0 & 2 & \textbf{9} & 6 \\
|
||
3 & 7 & 2 & 0 & 7 & 8 \\
|
||
4 & 9 & \textbf{9} & 7 & 0 & 2 \\
|
||
5 & 1 & 6 & 8 & 2 & 0 \\
|
||
\end{tabular}
|
||
\end{center}
|
||
\vspace{10pt}
|
||
|
||
|
||
Algoritmin toiminta jatkuu samalla tavalla
|
||
niin, että kukin solmu tulee vuorollaan
|
||
välisolmuksi.
|
||
Algoritmin päätteeksi taulukko sisältää
|
||
lyhimmän etäisyyden minkä tahansa
|
||
solmuparin välillä:
|
||
|
||
\begin{center}
|
||
\begin{tabular}{r|rrrrr}
|
||
& 1 & 2 & 3 & 4 & 5 \\
|
||
\hline
|
||
1 & 0 & 5 & 7 & 3 & 1 \\
|
||
2 & 5 & 0 & 2 & 9 & 6 \\
|
||
3 & 7 & 2 & 0 & 7 & 8 \\
|
||
4 & 3 & 9 & 7 & 0 & 2 \\
|
||
5 & 1 & 6 & 8 & 2 & 0 \\
|
||
\end{tabular}
|
||
\end{center}
|
||
|
||
Esimerkiksi taulukosta selviää, että lyhin polku
|
||
solmusta 2 solmuun 4 on pituudeltaan 8.
|
||
Tämä vastaa seuraavaa polkua:
|
||
|
||
\begin{center}
|
||
\begin{tikzpicture}[scale=0.9]
|
||
\node[draw, circle] (1) at (1,3) {$3$};
|
||
\node[draw, circle] (2) at (4,3) {$4$};
|
||
\node[draw, circle] (3) at (1,1) {$2$};
|
||
\node[draw, circle] (4) at (4,1) {$1$};
|
||
\node[draw, circle] (5) at (6,2) {$5$};
|
||
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=above:7] {} (2);
|
||
\path[draw,thick,-] (1) -- node[font=\small,label=left:2] {} (3);
|
||
\path[draw,thick,-] (3) -- node[font=\small,label=below:5] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=left:9] {} (4);
|
||
\path[draw,thick,-] (2) -- node[font=\small,label=above:2] {} (5);
|
||
\path[draw,thick,-] (4) -- node[font=\small,label=below:1] {} (5);
|
||
|
||
\path[draw=red,thick,->,line width=2pt] (3) -- (4);
|
||
\path[draw=red,thick,->,line width=2pt] (4) -- (5);
|
||
\path[draw=red,thick,->,line width=2pt] (5) -- (2);
|
||
\end{tikzpicture}
|
||
\end{center}
|
||
|
||
\subsubsection{Toteutus}
|
||
|
||
Floyd–Warshallin algoritmin etuna on,
|
||
että se on helppoa toteuttaa.
|
||
Seuraava toteutus muodostaa etäisyysmatriisin
|
||
\texttt{d}, jossa $\texttt{d}[a][b]$
|
||
on pienin etäisyys polulla solmusta $a$ solmuun $b$.
|
||
Aluksi algoritmi alustaa matriisin \texttt{d}
|
||
verkon vierusmatriisin \texttt{v} perusteella
|
||
(arvo $10^9$ kuvastaa ääretöntä):
|
||
|
||
\begin{lstlisting}
|
||
for (int i = 1; i <= n; i++) {
|
||
for (int j = 1; j <= n; j++) {
|
||
if (i == j) d[i][j] = 0;
|
||
else if (v[i][j]) d[i][j] = v[i][j];
|
||
else d[i][j] = 1e9;
|
||
}
|
||
}
|
||
\end{lstlisting}
|
||
|
||
Tämän jälkeen lyhimmät polut löytyvät seuraavasti:
|
||
|
||
\begin{lstlisting}
|
||
for (int k = 1; k <= n; k++) {
|
||
for (int i = 1; i <= n; i++) {
|
||
for (int j = 1; j <= n; j++) {
|
||
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
|
||
}
|
||
}
|
||
}
|
||
\end{lstlisting}
|
||
|
||
Algoritmin aikavaativuus on
|
||
$O(n^3)$, koska siinä on kolme sisäkkäistä
|
||
silmukkaa,
|
||
jotka käyvät läpi verkon solmut.
|
||
|
||
Koska Floyd–Warshallin
|
||
algoritmin toteutus on yksinkertainen,
|
||
algoritmi voi olla hyvä valinta jopa silloin,
|
||
kun haettavana on yksittäinen
|
||
lyhin polku verkossa.
|
||
Tämä on kuitenkin mahdollista vain silloin,
|
||
kun verkko on niin pieni,
|
||
että kuutiollinen aikavaativuus on riittävä.
|