Dijkstra's algorithm

Dijkstra's algorithm is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

This page uses content from Wikipedia. The original article was at Dijkstra's algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with nonnegative edge path costs, producing a shortest path tree. This algorithm is often used in routing and as a subroutine in other graph algorithms.

For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex. It can also be used for finding costs of shortest paths from a single vertex to a single destination vertex by stopping the algorithm once the shortest path to the destination vertex has been determined. For example, if the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities. As a result, the shortest path first is widely used in network routing protocols, most notably IS-IS and OSPF (Open Shortest Path First).

Task:

1. Implement Dijkstra's algorithm.
2. Document the specific algorithm implemented. The {{trans}} template is sufficient. Otherwise add text outside of your program or add comments within your program. This is not a requirement to explain how the algorithm works, but to state which algorithm is implemented. If your code follows an external source such as the Wikipedia pseudocode, you can state that. You can state if it is Dijkstra's original algorithm or some more efficient variant. It is relevant to mention things like priority queues, heaps, and expected time complexity in big-O notation. If a priority queue is used, it is important to discuss how the step of decreasing the distance of a node is accomplished, and whether it is linear or logarithmic time.
3. Run your program with the following graph to find the shortest path between vertex "a" and vertex "e."
4. Show the output of your program.

Vertices

Number	Name
1	a
2	b
3	c
4	d
5	e
6	f

Edges

End points	Cost
a, b	7
a, c	9
a, f	14
b, c	10
b, d	15
c, d	11
c, f	2
d, e	6
e, f	9

You can use numbers or names to identify vertices in your program.

C++

(Modified from LiteratePrograms, which is MIT/X11 licensed.)

Solution follows Dijkstra's algorithm as described elsewhere. Data like min-distance, previous node, neighbors, are kept in separate data structures instead of part of the vertex. The vertex is simply represented as a string.

For the priority queue of vertexes, we use a self-balancing binary search tree (std::set), which should bound time complexity by O(E log V). Although C++ has heaps, without knowing the index of an element it would take linear time to find it to re-order it for a changed weight. It is not easy to keep the index of vertexes in the heap because the heap operations are opaque without callbacks. On the other hand, using a self-balancing binary search tree is efficient because it has the same log(n) complexity for insertion and removal of the head element as a binary heap. In addition, a self-balancing binary search tree also allows us to find and remove any other element in log(n) time, allowing us to perform the decrease-key step in logarithmic time by removing and re-inserting.

We do not need to keep track of whether a vertex is "done" ("visited") as in the Wikipedia description, since re-reaching such a vertex will always fail the relaxation condition (when re-reaching a "done" vertex, the new distance will never be less than it was originally), so it will be skipped anyway.

<lang cpp>#include <iostream>

1. include <vector>
2. include <string>
3. include <map>
4. include <list>

1. include <limits> // for numeric_limits

1. include <set>
2. include <utility> // for pair
3. include <algorithm>
4. include <iterator>

typedef std::string vertex_t; typedef int weight_t;

const int max_weight = std::numeric_limits<int>::max();

struct neighbor {

   vertex_t target;
   weight_t weight;
   neighbor(vertex_t arg_target, weight_t arg_weight)
       : target(arg_target), weight(arg_weight) { }

};

typedef std::map<vertex_t, std::vector<neighbor> > adjacency_map_t;

void DijkstraComputePaths(vertex_t source, vertex_t target,

                         const adjacency_map_t &adjacency_map,
                         std::map<vertex_t, weight_t> &min_distance,
                         std::map<vertex_t, vertex_t> &previous)

{

   for (adjacency_map_t::const_iterator vertex_iter = adjacency_map.begin();
        vertex_iter != adjacency_map.end();
        vertex_iter++)
   {
       vertex_t v = vertex_iter->first;
       min_distance[v] = max_weight;
       for (std::vector<neighbor>::const_iterator neighbor_iter = vertex_iter->second.begin();
            neighbor_iter != vertex_iter->second.end();
            neighbor_iter++)
       {
           vertex_t v2 = neighbor_iter->target;
           min_distance[v2] = max_weight;
       }
   }
   min_distance[source] = 0;
   std::set<std::pair<weight_t, vertex_t> > vertex_queue;
   vertex_queue.insert(std::make_pair(min_distance[source], source));

   while (!vertex_queue.empty()) 
   {
       vertex_t u = vertex_queue.begin()->second;
       vertex_queue.erase(vertex_queue.begin());

if (u == target) break;

       // Visit each edge exiting u

const std::vector<neighbor> &neighbors = adjacency_map.find(u)->second;

       for (std::vector<neighbor>::const_iterator neighbor_iter = neighbors.begin();
            neighbor_iter != neighbors.end();
            neighbor_iter++)
       {
           vertex_t v = neighbor_iter->target;
           weight_t weight = neighbor_iter->weight;
           weight_t distance_through_u = min_distance[u] + weight;

if (distance_through_u < min_distance[v]) { vertex_queue.erase(std::make_pair(min_distance[v], v));

min_distance[v] = distance_through_u; previous[v] = u; vertex_queue.insert(std::make_pair(min_distance[v], v));

}

       }
   }

}

std::list<vertex_t> DijkstraGetShortestPathTo(

   vertex_t target, const std::map<vertex_t, vertex_t> &previous)

{

   std::list<vertex_t> path;
   std::map<vertex_t, vertex_t>::const_iterator prev;
   vertex_t vertex = target;
   path.push_front(vertex);
   while((prev = previous.find(vertex)) != previous.end())
   {
       vertex = prev->second;
       path.push_front(vertex);
   }
   return path;

}

int main() {

   // remember to insert edges both ways for an undirected graph
   adjacency_map_t adjacency_map;
   adjacency_map["a"].push_back(neighbor("b", 7));
   adjacency_map["a"].push_back(neighbor("c", 9));
   adjacency_map["a"].push_back(neighbor("f", 14));
   adjacency_map["b"].push_back(neighbor("a", 7));
   adjacency_map["b"].push_back(neighbor("c", 10));
   adjacency_map["b"].push_back(neighbor("d", 15));
   adjacency_map["c"].push_back(neighbor("a", 9));
   adjacency_map["c"].push_back(neighbor("b", 10));
   adjacency_map["c"].push_back(neighbor("d", 11));
   adjacency_map["c"].push_back(neighbor("f", 2));
   adjacency_map["d"].push_back(neighbor("b", 15));
   adjacency_map["d"].push_back(neighbor("c", 11));
   adjacency_map["d"].push_back(neighbor("e", 6));
   adjacency_map["e"].push_back(neighbor("d", 6));
   adjacency_map["e"].push_back(neighbor("f", 9));
   adjacency_map["f"].push_back(neighbor("a", 14));
   adjacency_map["f"].push_back(neighbor("c", 2));
   adjacency_map["f"].push_back(neighbor("e", 9));

   std::map<vertex_t, weight_t> min_distance;
   std::map<vertex_t, vertex_t> previous;
   DijkstraComputePaths("a", "e", adjacency_map, min_distance, previous);
   std::cout << "Distance from a to e: " << min_distance["e"] << std::endl;
   std::list<vertex_t> path = DijkstraGetShortestPathTo("e", previous);
   std::cout << "Path : ";
   std::copy(path.begin(), path.end(), std::ostream_iterator<vertex_t>(std::cout, " "));
   std::cout << std::endl;

   return 0;

}</lang>

Go

Algorithm is derived from Wikipedia section 1, titled "Algorithm," with nodes stored in a heap, which should bound time complexity by O(E log V). Decreasing the distance of a node is accomplished by removing it from the heap and then re-inserting it. Linear-time traversal to find the node in the heap is avoided by keeping the index of each node in the heap as a field of the node, and our heap operations ensure that this variable is updated appropriately whenever nodes are moved in the heap. Thus removal from the heap is accomplished in logarithmic time.

Comments in the code below refer to corresponding steps in that section of the WP page. The steps are reordered somewhat, and a significant variation in step 2 is that the unvisited set is not initially populated. Instead, the unvisited set is maintained as the "tentative" set, as illustrated on the WP page in the animated image showing robot motion planning.

The heap support comes from the Go standard library. The three operations used here are each documented to have O(log n) complexity. <lang go>package main

import (

   "container/heap"
   "fmt"
   "math"

)

// edge struct holds the bare data needed to define a graph type edge struct {

   vert1, vert2 string
   dist         int

}

func main() {

   // example data
   g := []edge{
       {"a", "b", 7},
       {"a", "c", 9},
       {"a", "f", 14},
       {"b", "c", 10},
       {"b", "d", 15},
       {"c", "d", 11},
       {"c", "f", 2},
       {"d", "e", 6},
       {"e", "f", 9},
   }
   path, dist, n := shortestPath("a", "e", g)
   fmt.Printf("Example with %d nodes, %d edges\n", n, len(g))
   fmt.Println("Shortest path: ", path)
   fmt.Println("Path distance: ", dist)

}

// node and neighbor structs hold data useful for the heap-optimized // Dijkstra's shortest path algorithm type node struct {

   vert string     // vertex name
   tent int        // tentative distance
   prev *node      // previous node in shortest path back to start
   done bool       // true when tent and prev represent shortest path
   nbs  []neighbor // edges from this vertex
   rx   int        // heap.Remove index

}

type neighbor struct {

   nd   *node // node corresponding to vertex
   dist int   // distance to this node (from whatever node references this)

}

const maxInt = math.MaxInt32

// shortestPath runs Dijkstra's algorithm. // Inputs are names of start and end vertices, and a graph in the form // of a simple list of edges. // Output is the path found as a list of vertex names, and for convenience, // the the total path distance and number of nodes in the graph. // Output is an empty list if there is no path from start to end. func shortestPath(start, end string, graph []edge) (p []string, d, n int) {

   // Setup for the algorithm constructs a linked representation
   // of the input graph, with fields needed by the algorithm.
   // In the process we take care of WP steps 1 and 2.
   all := make(map[string]*node)
   // one pass over graph to collect nodes
   for _, e := range graph {
       // WP step 1: initialize tentative distance to maxInt
       if all[e.vert1] == nil {
           all[e.vert1] = &node{vert: e.vert1, tent: maxInt}
       }
       if all[e.vert2] == nil {
           all[e.vert2] = &node{vert: e.vert2, tent: maxInt}
       }
   }
   current := all[start] // WP step 2
   last := all[end]
   if current == nil || last == nil {
       return // start or end vertex not in graph
   }
   current.tent = 0 // WP step 1
   // second pass to link neighbors
   for _, e := range graph {
       n1 := all[e.vert1]
       n2 := all[e.vert2]
       n1.nbs = append(n1.nbs, neighbor{n2, e.dist})
       n2.nbs = append(n2.nbs, neighbor{n1, e.dist})
   }
   // WP step 2
   var unvis ndList

   for {
       // WP step 5: check for end of path
       if current == last {
           for p = []string{end}; current.vert != start; {
               current = current.prev
               p = append(p, current.vert)
           }
           for i := (len(p) + 1) / 2; i > 0; i-- {
               p[i-1], p[len(p)-i] = p[len(p)-i], p[i-1]
           }
           return p, last.tent, len(all)
       }
       // WP step 3: update tentative distances to neighbors
       for _, nb := range current.nbs {
           if nd := nb.nd; !nd.done {
               if d := current.tent + nb.dist; d < nd.tent {
                   if nd.prev != nil {
                       heap.Remove(&unvis, nd.rx)
                   }
                   nd.tent = d
                   nd.prev = current
                   heap.Push(&unvis, nd)
               }
           }
       }
       // WP step 4: mark current node visited
       current.done = true
       // WP step 6: new current is node with smallest tentative distance
       current = heap.Pop(&unvis).(*node)
       if current == nil {
           break // no path exists
       }
   }
   return

}

// ndList implements container/heap type ndList []*node

func (n ndList) Len() int { return len(n) } func (n ndList) Less(i, j int) bool { return n[i].tent < n[j].tent } func (n ndList) Swap(i, j int) {

   n[i], n[j] = n[j], n[i]
   n[i].rx = i 
   n[j].rx = j

} func (n *ndList) Push(x interface{}) {

   nd := x.(*node)
   nd.rx = len(*n)
   *n = append(*n, nd)

} func (n *ndList) Pop() interface{} {

   s := *n
   if len(s) == 0 {
       return nil 
   }
   last := len(s) - 1
   r := s[last]
   *n = s[:last]
   return r

}</lang> Output:

Example with 6 nodes, 9 edges
Shortest path:  [a c f e]
Path distance:  20

Java

Notes for this solution:

• The number of nodes is fixed to less than 50
• At every iteration, the next minimum distance node found by linear traversal of all nodes, which is inefficient.

<lang java>import java.io.*; import java.util.*;

class Graph {

   private static final int MAXNODES = 50;
   private static final int INFINITY = Integer.MAX_VALUE;
   int n;
   int[][] weight = new int[MAXNODES][MAXNODES];
   int[] distance = new int[MAXNODES];
   int[] precede = new int[MAXNODES];

   /**
    * Find the shortest path across the graph using Dijkstra's algorithm.
    */
   void buildSpanningTree(int source, int destination) {

boolean[] visit = new boolean[MAXNODES];

for (int i=0 ; i<n ; i++) { distance[i] = INFINITY; precede[i] = INFINITY; } distance[source] = 0;

int current = source; while (current != destination) { int distcurr = distance[current]; int smalldist = INFINITY; int k = -1; visit[current] = true;

for (int i=0; i<n; i++) { if (visit[i]) continue;

int newdist = distcurr + weight[current][i]; if (newdist < distance[i]) { distance[i] = newdist; precede[i] = current; } if (distance[i] < smalldist) { smalldist = distance[i]; k = i; } } current = k; }

   }

   /**
    * Get the shortest path across a tree that has had its path weights
    * calculated.
    */
   int[] getShortestPath(int source, int destination) {

int i = destination; int finall = 0; int[] path = new int[MAXNODES];

path[finall] = destination; finall++; while (precede[i] != source) { i = precede[i]; path[finall] = i; finall++; } path[finall] = source;

int[] result = new int[finall+1]; System.arraycopy(path, 0, result, 0, finall+1); return result;

   }

   /**
    * Print the result.
    */
   void displayResult(int[] path) {

System.out.println("\nThe shortest path followed is : \n"); for (int i = path.length-1 ; i>0 ; i--) System.out.println("\t\t( " + path[i] + " ->" + path[i-1] + " ) with cost = " + weight[path[i]][path[i-1]]); System.out.println("For the Total Cost = " + distance[path[path.length-1]]);

   }

   /**
    * Prompt for a number.
    */
   int getNumber(String msg) {

int ne = 0; BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

try { System.out.print("\n" + msg + " : "); ne = Integer.parseInt(in.readLine()); } catch (Exception e) { System.out.println("I/O Error"); } return ne;

   }

   /**
    * Prompt for a tree, build and display a path across it.
    */
   void SPA() {

n = getNumber("Enter the number of nodes (Less than " + MAXNODES + ") in the matrix");

System.out.print("\nEnter the cost matrix : \n\n"); for (int i=0 ; i<n ; i++) for (int j=0 ; j<n ; j++) weight[i][j] = getNumber("Cost " + (i+1) + "--" + (j+1));

int s = getNumber("Enter the source node"); int d = getNumber("Enter the destination node");

buildSpanningTree(s, d); displayResult(getShortestPath(s, d));

   }

}

public class Dijkstra {

   public static void main(String args[]) {

Graph g = new Graph(); g.SPA();

   }

}</lang>

Mathematica

<lang Mathematica>bd = Graph[ { "a"\[UndirectedEdge] "b", "a"\[UndirectedEdge] "c", "b"\[UndirectedEdge] "c", "b"\[UndirectedEdge] "d", "c"\[UndirectedEdge] "d", "d"\[UndirectedEdge] "e", "a"\[UndirectedEdge] "f", "c"\[UndirectedEdge] "f", "e"\[UndirectedEdge] "f" } , EdgeWeight->{7,9,10,15,11,6,14,2,9},VertexLabels->"Name", VertexLabelStyle->Directive[Black,20],ImagePadding->20]

FindShortestPath[bd, "a", "e", Method -> "Dijkstra"] -> {"a", "c", "f", "e"}</lang>

PARI/GP

Basic, inefficient implementation. Takes an n×n matrix representing distance between nodes (a 0-1 matrix if you just want to count number of steps) and a number in 1..n representing the starting node, which defaults to 1 if not given. <lang parigp>shortestPath(G, startAt=1)={ my(n=#G[,1],dist=vector(n,i,9e99),prev=dist,Q=2^n-1); dist[startAt]=0; while(Q, my(t=vecmin(vecextract(dist,Q)),u); if(t==9e99, break); for(i=1,#v,if(dist[i]==t && bittest(Q,i-1), u=i; break)); Q-=1<<(u-1); for(i=1,n, if(!G[u,i],next); my(alt=dist[u]+G[u,i]); if (alt < dist[i], dist[i]=alt; prev[i]=u; ) ) ); dist };</lang>

Ruby

Works with: Ruby version 1.9.2+

(for INFINITY)

Notes for this solution:

• At every iteration, the next minimum distance node found by linear traversal of all nodes, which is inefficient.

<lang ruby>class Graph

 Vertex = Struct.new(:name, :neighbours, :dist, :prev)
 Edge = Struct.new(:v1, :v2, :distance)
 class Edge
   def vertices; [v1, v2]; end
 end

 def initialize(graph)
   @vertices = {}
   @edges = []
   graph.each do |(v1, v2, dist)|
     @vertices[v1] = Vertex.new(v1, []) unless @vertices.has_key?(v1) 
     vert1 = @vertices[v1]
     @vertices[v2] = Vertex.new(v2, []) unless @vertices.has_key?(v2) 
     vert2 = @vertices[v2]
     vert1.neighbours << vert2
     vert2.neighbours << vert1
     @edges << Edge.new(vert1, vert2, dist)
   end
   @dijkstra_source = nil
 end
 attr_reader :vertices, :edges

 def edge(u, v)
   @edges.find {|e| e.vertices == [u, v] or e.vertices == [v, u]}
 end

 def dijkstra(source)
   q = vertices.values
   q.each {|v| v.dist = Float::INFINITY}
   source.dist = 0
   until q.empty?
     u = q.min_by {|vertex| vertex.dist}
     break if u == Float::INFINITY
     q.delete(u)
     u.neighbours.each do |v|
       if q.include?(v)
         alt = u.dist + edge(u,v).distance
         if alt < v.dist
           v.dist = alt
           v.prev = u
         end
       end
     end
   end
   @dijkstra_source = source
 end

 def shortest_path(source, target)
   dijkstra(source) unless @dijkstra_source == source
   path = []
   u = target
   until u.nil?
     path.unshift(u)
     u = u.prev
   end
   path
 end

 def to_s
   "#<%s vertices=%s edges=%s>" % [self.class.name, vertices.values.inspect, edges.inspect] 
 end

end

g = Graph.new([ [:a, :b, 7],

               [:a, :c, 9],
               [:b, :c, 10],
               [:b, :d, 15],
               [:c, :d, 11],
               [:d, :e, 6],
               [:a, :f, 14],
               [:c, :f, 2],
               [:e, :f, 9],
             ])

start = g.vertices[:a] stop = g.vertices[:e] path = g.shortest_path(start, stop) puts "shortest path from #{start.name} to #{stop.name} has cost #{stop.dist}:" puts path.map {|vertex| vertex.name}.join(" -> ")</lang>

output

shortest path from a to e has cost 20:
a -> c -> f -> e

Tcl

Note that this code traverses the entire set of unrouted nodes at each step, as this is simpler than computing the subset that are reachable at each stage. <lang tcl>proc dijkstra {graph origin} {

   # Initialize
   dict for {vertex distmap} $graph {

dict set dist $vertex Inf dict set path $vertex {}

   }
   dict set dist $origin 0
   dict set path $origin [list $origin]

   while {[dict size $graph]} {

# Find unhandled node with least weight set d Inf dict for {uu -} $graph { if {$d > [set dd [dict get $dist $uu]]} { set u $uu set d $dd } }

# No such node; graph must be disconnected if {$d == Inf} break

# Update the weights for nodes lead to by the node we've picked dict for {v dd} [dict get $graph $u] { if {[dict exists $graph $v]} { set alt [expr {$d + $dd}] if {$alt < [dict get $dist $v]} { dict set dist $v $alt dict set path $v [list {*}[dict get $path $u] $v] } } }

# Remove chosen node from graph still to be handled dict unset graph $u

   }
   return [list $dist $path]

}</lang> Showing the code in use: <lang tcl>proc makeUndirectedGraph arcs {

   # Assume that all nodes are connected to something
   foreach arc $arcs {

lassign $arc v1 v2 cost dict set graph $v1 $v2 $cost dict set graph $v2 $v1 $cost

   }
   return $graph

} set arcs {

   {a b 7} {a c 9} {b c 10} {b d 15} {c d 11}
   {d e 6} {a f 14} {c f 2} {e f 9}

} lassign [dijkstra [makeUndirectedGraph $arcs] "a"] costs path puts "path from a to e costs [dict get $costs e]" puts "route from a to e is: [join [dict get $path e] { -> }]"</lang> Output:

path from a to e costs 20
route from a to e is: a -> c -> f -> e