Floyd’s Algorithm

Akash

Lesson Progress

0% Complete

Floyd-Warshall Algorithm is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the edges in a cycle is negative).

A weighted graph is a graph in which each edge has a numerical value associated with it.

Floyd-Warhshall algorithm is also called as Floyd’s algorithm, Roy-Floyd algorithm, Roy-Warshall algorithm, or WFI algorithm.

This algorithm follows the dynamic programming approach to find the shortest paths.

Example:

Input:
       graph[][] = { {0,   5,  INF, 10},
                    {INF,  0,  3,  INF},
                    {INF, INF, 0,   1},
                    {INF, INF, INF, 0} }
which represents the following graph
             10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/         |
       (1)------->(2)
            3       
Note that the value of graph[i][j] is 0 if i is equal to j 
And graph[i][j] is INF (infinite) if there is no edge from vertex i to j.

Output:
Shortest distance matrix
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

Floyd-Warshall Algorithm

n = no of vertices
A = matrix of dimension n*n
for k = 1 to n
    for i = 1 to n
        for j = 1 to n
            A^k[i, j] = min (A^k-1[i, j], A^k-1[i, k] + A^k-1[k, j])
return A

Implementation of Floyd’s Algorithm

// C Program for Floyd Warshall Algorithm 
#include<stdio.h> 
  
// Number of vertices in the graph 
#define V 4 
  
/* Define Infinite as a large enough value. This value will be used 
  for vertices not connected to each other */
#define INF 99999 
  
// A function to print the solution matrix 
void printSolution(int dist[][V]); 
  
// Solves the all-pairs shortest path problem using Floyd Warshall algorithm 
void floydWarshall (int graph[][V]) 
{ 
    /* dist[][] will be the output matrix that will finally have the shortest  
      distances between every pair of vertices */
    int dist[V][V], i, j, k; 
  
    /* Initialize the solution matrix same as input graph matrix. Or  
       we can say the initial values of shortest distances are based 
       on shortest paths considering no intermediate vertex. */
    for (i = 0; i < V; i++) 
        for (j = 0; j < V; j++) 
            dist[i][j] = graph[i][j]; 
  
    /* Add all vertices one by one to the set of intermediate vertices. 
      ---> Before start of an iteration, we have shortest distances between all 
      pairs of vertices such that the shortest distances consider only the 
      vertices in set {0, 1, 2, .. k-1} as intermediate vertices. 
      ----> After the end of an iteration, vertex no. k is added to the set of 
      intermediate vertices and the set becomes {0, 1, 2, .. k} */
    for (k = 0; k < V; k++) 
    { 
        // Pick all vertices as source one by one 
        for (i = 0; i < V; i++) 
        { 
            // Pick all vertices as destination for the 
            // above picked source 
            for (j = 0; j < V; j++) 
            { 
                // If vertex k is on the shortest path from 
                // i to j, then update the value of dist[i][j] 
                if (dist[i][k] + dist[k][j] < dist[i][j]) 
                    dist[i][j] = dist[i][k] + dist[k][j]; 
            } 
        } 
    } 
  
    // Print the shortest distance matrix 
    printSolution(dist); 
} 
  
/* A utility function to print solution */
void printSolution(int dist[][V]) 
{ 
    printf ("The following matrix shows the shortest distances"
            " between every pair of vertices \n"); 
    for (int i = 0; i < V; i++) 
    { 
        for (int j = 0; j < V; j++) 
        { 
            if (dist[i][j] == INF) 
                printf("%7s", "INF"); 
            else
                printf ("%7d", dist[i][j]); 
        } 
        printf("\n"); 
    } 
} 
  
// driver program to test above function 
int main() 
{ 
    /* Let us create the following weighted graph 
            10 
       (0)------->(3) 
        |         /|\ 
      5 |          | 
        |          | 1 
       \|/         | 
       (1)------->(2) 
            3           */
    int graph[V][V] = { {0,   5,  INF, 10}, 
                        {INF, 0,   3, INF}, 
                        {INF, INF, 0,   1}, 
                        {INF, INF, INF, 0} 
                      }; 
  
    // Print the solution 
    floydWarshall(graph); 
    return 0; 
}

Time Complexity

Following matrix shows the shortest distances between every pair of vertices
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

Floyd Warshall Algorithm Complexity

Time Complexity

There are three loops. Each loop has constant complexities. So, the time complexity of the Floyd-Warshall algorithm is O(n³).

Space Complexity

The space complexity of the Floyd-Warshall algorithm is O(n²).

Floyd Warshall Algorithm Applications

To find the shortest path is a directed graph
To find the transitive closure of directed graphs
To find the Inversion of real matrices
For testing whether an undirected graph is bipartite

Algorithm

Participants2253