Dijkstra's Algorithm

Dijkstra's Algorithm

Dijkstra's algorithm allows us to find the shortest path between any two vertices of a graph.

It differs from the minimum spanning tree because the shortest distance between two vertices might not include all the vertices of the graph.


How Dijkstra's Algorithm works

Dijkstra's Algorithm works on the basis that any subpath B -> D of the shortest path A -> D between vertices A and D is also the shortest path between vertices B and D.

shortest subpath property is used by dijkstra's algorithm
Each subpath is the shortest path

Djikstra used this property in the opposite direction i.e we overestimate the distance of each vertex from the starting vertex. Then we visit each node and its neighbors to find the shortest subpath to those neighbors.

The algorithm uses a greedy approach in the sense that we find the next best solution hoping that the end result is the best solution for the whole problem.


Example of Dijkstra's algorithm

It is easier to start with an example and then think about the algorithm.

Start with a weighted graph
Start with a weighted graph
Choose a starting vertex and assign infinity path values to all other devices
Choose a starting vertex and assign infinity path values to all other devices
Go to each vertex and update its path length
Go to each vertex and update its path length
If the path length of the adjacent vertex is lesser than new path length, don't update it
If the path length of the adjacent vertex is lesser than new path length, don't update it
Avoid updating path lengths of already visited vertices
Avoid updating path lengths of already visited vertices
After each iteration, we pick the unvisited vertex with the least path length. So we choose 5 before 7
After each iteration, we pick the unvisited vertex with the least path length. So we choose 5 before 7
Notice how the rightmost vertex has its path length updated twice
Notice how the rightmost vertex has its path length updated twice
Repeat until all the vertices have been visited
Repeat until all the vertices have been visited

Djikstra's algorithm pseudocode

We need to maintain the path distance of every vertex. We can store that in an array of size v, where v is the number of vertices.

We also want to be able to get the shortest path, not only know the length of the shortest path. For this, we map each vertex to the vertex that last updated its path length.

Once the algorithm is over, we can backtrack from the destination vertex to the source vertex to find the path.

A minimum priority queue can be used to efficiently receive the vertex with least path distance.

function dijkstra(G, S)
    for each vertex V in G
        distance[V] <- infinite
        previous[V] <- NULL
        If V != S, add V to Priority Queue Q
    distance[S] <- 0
	
    while Q IS NOT EMPTY
        U <- Extract MIN from Q
        for each unvisited neighbour V of U
            tempDistance <- distance[U] + edge_weight(U, V)
            if tempDistance < distance[V]
                distance[V] <- tempDistance
                previous[V] <- U
    return distance[], previous[]

Code for Dijkstra's Algorithm

The implementation of Dijkstra's Algorithm in C++ is given below. The complexity of the code can be improved, but the abstractions are convenient to relate the code with the algorithm.

# Dijkstra's Algorithm in Python


import sys

# Providing the graph
vertices = [[0, 0, 1, 1, 0, 0, 0],
            [0, 0, 1, 0, 0, 1, 0],
            [1, 1, 0, 1, 1, 0, 0],
            [1, 0, 1, 0, 0, 0, 1],
            [0, 0, 1, 0, 0, 1, 0],
            [0, 1, 0, 0, 1, 0, 1],
            [0, 0, 0, 1, 0, 1, 0]]

edges = [[0, 0, 1, 2, 0, 0, 0],
         [0, 0, 2, 0, 0, 3, 0],
         [1, 2, 0, 1, 3, 0, 0],
         [2, 0, 1, 0, 0, 0, 1],
         [0, 0, 3, 0, 0, 2, 0],
         [0, 3, 0, 0, 2, 0, 1],
         [0, 0, 0, 1, 0, 1, 0]]

# Find which vertex is to be visited next
def to_be_visited():
    global visited_and_distance
    v = -10
    for index in range(num_of_vertices):
        if visited_and_distance[index][0] == 0 \
            and (v < 0 or visited_and_distance[index][1] <=
                 visited_and_distance[v][1]):
            v = index
    return v


num_of_vertices = len(vertices[0])

visited_and_distance = [[0, 0]]
for i in range(num_of_vertices-1):
    visited_and_distance.append([0, sys.maxsize])

for vertex in range(num_of_vertices):

    # Find next vertex to be visited
    to_visit = to_be_visited()
    for neighbor_index in range(num_of_vertices):

        # Updating new distances
        if vertices[to_visit][neighbor_index] == 1 and \
                visited_and_distance[neighbor_index][0] == 0:
            new_distance = visited_and_distance[to_visit][1] \
                + edges[to_visit][neighbor_index]
            if visited_and_distance[neighbor_index][1] > new_distance:
                visited_and_distance[neighbor_index][1] = new_distance
        
        visited_and_distance[to_visit][0] = 1

i = 0

# Printing the distance
for distance in visited_and_distance:
    print("Distance of ", chr(ord('a') + i),
          " from source vertex: ", distance[1])
    i = i + 1
// Dijkstra's Algorithm in Java

public class Dijkstra {

  public static void dijkstra(int[][] graph, int source) {
    int count = graph.length;
    boolean[] visitedVertex = new boolean[count];
    int[] distance = new int[count];
    for (int i = 0; i < count; i++) {
      visitedVertex[i] = false;
      distance[i] = Integer.MAX_VALUE;
    }

    // Distance of self loop is zero
    distance[source] = 0;
    for (int i = 0; i < count; i++) {

      // Update the distance between neighbouring vertex and source vertex
      int u = findMinDistance(distance, visitedVertex);
      visitedVertex[u] = true;

      // Update all the neighbouring vertex distances
      for (int v = 0; v < count; v++) {
        if (!visitedVertex[v] && graph[u][v] != 0 && (distance[u] + graph[u][v] < distance[v])) {
          distance[v] = distance[u] + graph[u][v];
        }
      }
    }
    for (int i = 0; i < distance.length; i++) {
      System.out.println(String.format("Distance from %s to %s is %s", source, i, distance[i]));
    }

  }

  // Finding the minimum distance
  private static int findMinDistance(int[] distance, boolean[] visitedVertex) {
    int minDistance = Integer.MAX_VALUE;
    int minDistanceVertex = -1;
    for (int i = 0; i < distance.length; i++) {
      if (!visitedVertex[i] && distance[i] < minDistance) {
        minDistance = distance[i];
        minDistanceVertex = i;
      }
    }
    return minDistanceVertex;
  }

  public static void main(String[] args) {
    int graph[][] = new int[][] { { 0, 0, 1, 2, 0, 0, 0 }, { 0, 0, 2, 0, 0, 3, 0 }, { 1, 2, 0, 1, 3, 0, 0 },
        { 2, 0, 1, 0, 0, 0, 1 }, { 0, 0, 3, 0, 0, 2, 0 }, { 0, 3, 0, 0, 2, 0, 1 }, { 0, 0, 0, 1, 0, 1, 0 } };
    Dijkstra T = new Dijkstra();
    T.dijkstra(graph, 0);
  }
}
// Dijkstra's Algorithm in C

#include <stdio.h>
#define INFINITY 9999
#define MAX 10

void Dijkstra(int Graph[MAX][MAX], int n, int start);

void Dijkstra(int Graph[MAX][MAX], int n, int start) {
  int cost[MAX][MAX], distance[MAX], pred[MAX];
  int visited[MAX], count, mindistance, nextnode, i, j;

  // Creating cost matrix
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      if (Graph[i][j] == 0)
        cost[i][j] = INFINITY;
      else
        cost[i][j] = Graph[i][j];

  for (i = 0; i < n; i++) {
    distance[i] = cost[start][i];
    pred[i] = start;
    visited[i] = 0;
  }

  distance[start] = 0;
  visited[start] = 1;
  count = 1;

  while (count < n - 1) {
    mindistance = INFINITY;

    for (i = 0; i < n; i++)
      if (distance[i] < mindistance && !visited[i]) {
        mindistance = distance[i];
        nextnode = i;
      }

    visited[nextnode] = 1;
    for (i = 0; i < n; i++)
      if (!visited[i])
        if (mindistance + cost[nextnode][i] < distance[i]) {
          distance[i] = mindistance + cost[nextnode][i];
          pred[i] = nextnode;
        }
    count++;
  }

  // Printing the distance
  for (i = 0; i < n; i++)
    if (i != start) {
      printf("\nDistance from source to %d: %d", i, distance[i]);
    }
}
int main() {
  int Graph[MAX][MAX], i, j, n, u;
  n = 7;

  Graph[0][0] = 0;
  Graph[0][1] = 0;
  Graph[0][2] = 1;
  Graph[0][3] = 2;
  Graph[0][4] = 0;
  Graph[0][5] = 0;
  Graph[0][6] = 0;

  Graph[1][0] = 0;
  Graph[1][1] = 0;
  Graph[1][2] = 2;
  Graph[1][3] = 0;
  Graph[1][4] = 0;
  Graph[1][5] = 3;
  Graph[1][6] = 0;

  Graph[2][0] = 1;
  Graph[2][1] = 2;
  Graph[2][2] = 0;
  Graph[2][3] = 1;
  Graph[2][4] = 3;
  Graph[2][5] = 0;
  Graph[2][6] = 0;

  Graph[3][0] = 2;
  Graph[3][1] = 0;
  Graph[3][2] = 1;
  Graph[3][3] = 0;
  Graph[3][4] = 0;
  Graph[3][5] = 0;
  Graph[3][6] = 1;

  Graph[4][0] = 0;
  Graph[4][1] = 0;
  Graph[4][2] = 3;
  Graph[4][3] = 0;
  Graph[4][4] = 0;
  Graph[4][5] = 2;
  Graph[4][6] = 0;

  Graph[5][0] = 0;
  Graph[5][1] = 3;
  Graph[5][2] = 0;
  Graph[5][3] = 0;
  Graph[5][4] = 2;
  Graph[5][5] = 0;
  Graph[5][6] = 1;

  Graph[6][0] = 0;
  Graph[6][1] = 0;
  Graph[6][2] = 0;
  Graph[6][3] = 1;
  Graph[6][4] = 0;
  Graph[6][5] = 1;
  Graph[6][6] = 0;

  u = 0;
  Dijkstra(Graph, n, u);

  return 0;
}
// Dijkstra's Algorithm in C++

#include <iostream>
#include <vector>

#define INT_MAX 10000000

using namespace std;

void DijkstrasTest();

int main() {
  DijkstrasTest();
  return 0;
}

class Node;
class Edge;

void Dijkstras();
vector<Node*>* AdjacentRemainingNodes(Node* node);
Node* ExtractSmallest(vector<Node*>& nodes);
int Distance(Node* node1, Node* node2);
bool Contains(vector<Node*>& nodes, Node* node);
void PrintShortestRouteTo(Node* destination);

vector<Node*> nodes;
vector<Edge*> edges;

class Node {
   public:
  Node(char id)
    : id(id), previous(NULL), distanceFromStart(INT_MAX) {
    nodes.push_back(this);
  }

   public:
  char id;
  Node* previous;
  int distanceFromStart;
};

class Edge {
   public:
  Edge(Node* node1, Node* node2, int distance)
    : node1(node1), node2(node2), distance(distance) {
    edges.push_back(this);
  }
  bool Connects(Node* node1, Node* node2) {
    return (
      (node1 == this->node1 &&
       node2 == this->node2) ||
      (node1 == this->node2 &&
       node2 == this->node1));
  }

   public:
  Node* node1;
  Node* node2;
  int distance;
};

///////////////////
void DijkstrasTest() {
  Node* a = new Node('a');
  Node* b = new Node('b');
  Node* c = new Node('c');
  Node* d = new Node('d');
  Node* e = new Node('e');
  Node* f = new Node('f');
  Node* g = new Node('g');

  Edge* e1 = new Edge(a, c, 1);
  Edge* e2 = new Edge(a, d, 2);
  Edge* e3 = new Edge(b, c, 2);
  Edge* e4 = new Edge(c, d, 1);
  Edge* e5 = new Edge(b, f, 3);
  Edge* e6 = new Edge(c, e, 3);
  Edge* e7 = new Edge(e, f, 2);
  Edge* e8 = new Edge(d, g, 1);
  Edge* e9 = new Edge(g, f, 1);

  a->distanceFromStart = 0;  // set start node
  Dijkstras();
  PrintShortestRouteTo(f);
}

///////////////////

void Dijkstras() {
  while (nodes.size() > 0) {
    Node* smallest = ExtractSmallest(nodes);
    vector<Node*>* adjacentNodes =
      AdjacentRemainingNodes(smallest);

    const int size = adjacentNodes->size();
    for (int i = 0; i < size; ++i) {
      Node* adjacent = adjacentNodes->at(i);
      int distance = Distance(smallest, adjacent) +
               smallest->distanceFromStart;

      if (distance < adjacent->distanceFromStart) {
        adjacent->distanceFromStart = distance;
        adjacent->previous = smallest;
      }
    }
    delete adjacentNodes;
  }
}

// Find the node with the smallest distance,
// remove it, and return it.
Node* ExtractSmallest(vector<Node*>& nodes) {
  int size = nodes.size();
  if (size == 0) return NULL;
  int smallestPosition = 0;
  Node* smallest = nodes.at(0);
  for (int i = 1; i < size; ++i) {
    Node* current = nodes.at(i);
    if (current->distanceFromStart <
      smallest->distanceFromStart) {
      smallest = current;
      smallestPosition = i;
    }
  }
  nodes.erase(nodes.begin() + smallestPosition);
  return smallest;
}

// Return all nodes adjacent to 'node' which are still
// in the 'nodes' collection.
vector<Node*>* AdjacentRemainingNodes(Node* node) {
  vector<Node*>* adjacentNodes = new vector<Node*>();
  const int size = edges.size();
  for (int i = 0; i < size; ++i) {
    Edge* edge = edges.at(i);
    Node* adjacent = NULL;
    if (edge->node1 == node) {
      adjacent = edge->node2;
    } else if (edge->node2 == node) {
      adjacent = edge->node1;
    }
    if (adjacent && Contains(nodes, adjacent)) {
      adjacentNodes->push_back(adjacent);
    }
  }
  return adjacentNodes;
}

// Return distance between two connected nodes
int Distance(Node* node1, Node* node2) {
  const int size = edges.size();
  for (int i = 0; i < size; ++i) {
    Edge* edge = edges.at(i);
    if (edge->Connects(node1, node2)) {
      return edge->distance;
    }
  }
  return -1;  // should never happen
}

// Does the 'nodes' vector contain 'node'
bool Contains(vector<Node*>& nodes, Node* node) {
  const int size = nodes.size();
  for (int i = 0; i < size; ++i) {
    if (node == nodes.at(i)) {
      return true;
    }
  }
  return false;
}

///////////////////

void PrintShortestRouteTo(Node* destination) {
  Node* previous = destination;
  cout << "Distance from start: "
     << destination->distanceFromStart << endl;
  while (previous) {
    cout << previous->id << " ";
    previous = previous->previous;
  }
  cout << endl;
}

// these two not needed
vector<Edge*>* AdjacentEdges(vector<Edge*>& Edges, Node* node);
void RemoveEdge(vector<Edge*>& Edges, Edge* edge);

vector<Edge*>* AdjacentEdges(vector<Edge*>& edges, Node* node) {
  vector<Edge*>* adjacentEdges = new vector<Edge*>();

  const int size = edges.size();
  for (int i = 0; i < size; ++i) {
    Edge* edge = edges.at(i);
    if (edge->node1 == node) {
      cout << "adjacent: " << edge->node2->id << endl;
      adjacentEdges->push_back(edge);
    } else if (edge->node2 == node) {
      cout << "adjacent: " << edge->node1->id << endl;
      adjacentEdges->push_back(edge);
    }
  }
  return adjacentEdges;
}

void RemoveEdge(vector<Edge*>& edges, Edge* edge) {
  vector<Edge*>::iterator it;
  for (it = edges.begin(); it < edges.end(); ++it) {
    if (*it == edge) {
      edges.erase(it);
      return;
    }
  }
}

Dijkstra's Algorithm Complexity

Time Complexity: O(E Log V)

where, E is the number of edges and V is the number of vertices.

Space Complexity: O(V)


Dijkstra's Algorithm Applications

  • To find the shortest path
  • In social networking applications
  • In a telephone network
  • To find the locations in the map