Dijkstra’s algorithm was introduced by Dutch computer scientist Edsger W. Dijkstra in 1956. It is an iterative algorithm for finding shortest path between nodes in a data structure called graph. Graph data structure consists of a finite set of nodes and edges. The objective of introducing this algorithm was to determine the capabilities of then developed computer “ARMAC”. Since, graph can also be represented as road networks. So, Dijkstra in 1956 designed this shortest path algorithm and implemented it for ARMAC for transportation map of 64 cities in Netherland.
The following conditions should be fulfilled before applying Dijkstra’s algorithm:
To make the implementation easy, we’ll take the Priority Queues for storing unvisited node. This will make the extraction of the node on the basis of minimum cost easy.
dijkstra(G, S) dist[source] <- 0 previous[source] <- NULL for each node N in G if N != S dist[N] <- infinite previous[N] <- NULL ADD V to Q while Q IS NOT EMPTY U <- Extract MIN from Q for each unvisited neighbour V of U tempDist <- dist[U] + edge_weight(U, V) if tempDist < dist[V] dist[V] <- tempDist previous[V] <- U return dist[], previous[]
The program given below is for adjacency matrix representation of the graph
import java.util.*; import java.lang.*; import java.io.*; class Dijkstra { static final int N=9; // list will contain the set of unvisited nodes // this function will return the node with minimum distance value int minVal(int dist[], Boolean list[]) { int min = Integer.MAX_VALUE; int min_index=-1; for (int i= 0; i < V; i++) if (list[i] == false && dist[i] <= min) { min = dist[i]; min_index = i; } return min_index; } // print distance array void print(int dist[], int n) { System.out.println("Distance from source"); for (int j = 0; j< N; j++) System.out.println(j+" \t\t "+dist[j]); } void dijkstra(int graph[][], int src) { //dist[i] will hold the distance from src to i int dist[] = new int[N]; // list[i] will be true when the minimum distance will be achieved Boolean list[] = new Boolean[N]; for (int i = 0; i < N; i++) { dist[i] = Integer.MAX_VALUE; list[i] = false; } dist[src] = 0; for (int c = 0; c < N-1; c++) { // node of minimum distance value will be selected to be as a source. int u = minVal(dist, list); list[u] = true; // Update dist value of the adjacent nodes of the current node for (int v = 0; v < N; v++) if (!list[v] && graph[u][v]!=0 && dist[u] != Integer.MAX_VALUE && dist[u]+graph[u][v] < dist[v]) dist[v] = dist[u] + graph[u][v]; } // print the constructed distance array print(dist, N); } public static void main (String[] args) { // create the example graph int graph[][] = new int[][]{ {0, 4, 0, 0, 0, 0, 0, 8, 0}, {4, 0, 8, 0, 0, 0, 0, 11, 0}, {0, 8, 0, 7, 0, 4, 0, 0, 2}, {0, 0, 7, 0, 9, 14, 0, 0, 0}, {0, 0, 0, 9, 0, 10, 0, 0, 0}, {0, 0, 4, 14, 10, 0, 2, 0, 0}, {0, 0, 0, 0, 0, 2, 0, 1, 6}, {8, 11, 0, 0, 0, 0, 1, 0, 7}, {0, 0, 2, 0, 0, 0, 6, 7, 0} }; ShortestPath s= new ShortestPath(); s.dijkstra(graph, 0); } }
The program given below is for adjacency matrix representation of the graph
#include < stdio.h> #include < limits.h> // total nodes in the graph #define N 9 // list will contain the set of unvisited nodes // this function will return the node with minimum distance value int minDistance(int dist[], bool list[]) { int min = INT_MAX; int min_index; for (int i = 0; i < N; i++) if (list[i] == false && dist[i] <= min) min = dist[i]; min_index = v; return min_index; } // print distance array int print(int dist[], int n) { int i=0; printf("Distance of node from source\n"); for (i = 0; i < N; i++) printf("%d \t\t %d\n", i, dist[i]); } void dijkstra(int graph[N][N], int src) { int i, j, count; int dist[N]; //dist[i] will hold the distance from src to i bool list[N]; // list[i] will be true when the minimum distance will be achieved // Initialize all distances as INFINITE and list[] as false for ( i = 0; i < N; i++) dist[i] = INT_MAX; list[i] = false; dist[src] = 0; // Find shortest path for all nodes for ( count = 0; count < V-1; count++) { // node of minimum distance value will be selected to be as a source int u = minDistance(dist, list); // Mark the picked node as unvisited list[u] = true; // Update dist value of the adjacent vertices of the picked vertex. for (j = 0; j< N; j++) if (!list [j] && graph[u][j] && dist[j] != INT_MAX && dist[u]+graph[u][j] < dist[j]) dist[j] = dist[u] + graph[u][j]; } // print the constructed distance array print(dist, N); } int main() { // create the example graph int graph[V][V] = { {0, 4, 0, 0, 0, 0, 0, 8, 0}, {4, 0, 8, 0, 0, 0, 0, 11, 0}, {0, 8, 0, 7, 0, 4, 0, 0, 2}, {0, 0, 7, 0, 9, 14, 0, 0, 0}, {0, 0, 0, 9, 0, 10, 0, 0, 0}, {0, 0, 4, 14, 10, 0, 2, 0, 0}, {0, 0, 0, 0, 0, 2, 0, 1, 6}, {8, 11, 0, 0, 0, 0, 1, 0, 7}, {0, 0, 2, 0, 0, 0, 6, 7, 0} }; dijkstra(graph, 0); return 0; }
The list which we used for storing unvisited node, with different implementations produces different time complexity.
List as a linear array : If there are N nodes then, min_distance function takes O(N) time and number of operations is |N|. Therefore, in this function, it will take O(N^2) time.
The total number of edges let’s assume is E. That means for loop will iterate |E| times. Hence, the running time of this algorithm using linear array implementation is O(N2 + E) = O(N2).
List as a binary heap: If there are N number of operations and to build a binary heap. It will take O(N) time. Then, min_distance function will take O(log N) time.
If the number of edges is |E| and the graph is sparse. Then the total running time with binary heap implementation is O(E log N).
Fibonacci heap is the more complicated version of priority queue implementation. It implements increase_priority in O(1) time. If the number of edges is |E| and there are N number of nodes.
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So, the time complexity of Dijkstra’s algorithm with Fibonacci heap implementation will be O(N log N+ E).
When there are large constant factors involved. Then, in that case, the Fibonacci heap implementation becomes impractical.
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