Floyd Warshall Algorithm | DP-16

The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. The problem is to find shortest distances between every pair of vertices in a given edge weighted directed Graph.

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Floyd Warshall Algorithm
We initialize the solution matrix same as the input graph matrix as a first step. Then we update the solution matrix by considering all vertices as an intermediate vertex. The idea is to one by one pick all vertices and updates all shortest paths which include the picked vertex as an intermediate vertex in the shortest path. When we pick vertex number k as an intermediate vertex, we already have considered vertices {0, 1, 2, .. k-1} as intermediate vertices. For every pair (i, j) of the source and destination vertices respectively, there are two possible cases.
1) k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is.
2) k is an intermediate vertex in shortest path from i to j. We update the value of dist[i][j] as dist[i][k] + dist[k][j] if dist[i][j] > dist[i][k] + dist[k][j]

The following figure shows the above optimal substructure property in the all-pairs shortest path problem.

Following is implementations of the Floyd Warshall algorithm.

C/C++

// 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; 
} 

Java

// A Java program for Floyd Warshall All Pairs Shortest 
// Path algorithm. 
import java.util.*; 
import java.lang.*; 
import java.io.*; 
  
  
class AllPairShortestPath 
{ 
    final static int INF = 99999, V = 4; 
  
    void floydWarshall(int graph[][]) 
    { 
        int dist[][] = new int[V][V]; 
        int 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); 
    } 
  
    void printSolution(int dist[][]) 
    { 
        System.out.println("The following matrix shows the shortest "+ 
                         "distances between every pair of vertices"); 
        for (int i=0; i<V; ++i) 
        { 
            for (int j=0; j<V; ++j) 
            { 
                if (dist[i][j]==INF) 
                    System.out.print("INF "); 
                else
                    System.out.print(dist[i][j]+"   "); 
            } 
            System.out.println(); 
        } 
    } 
  
    // Driver program to test above function 
    public static void main (String[] args) 
    { 
        /* Let us create the following weighted graph 
           10 
        (0)------->(3) 
        |         /|\ 
        5 |          | 
        |          | 1 
        \|/         | 
        (1)------->(2) 
           3           */
        int graph[][] = { {0,   5,  INF, 10}, 
                          {INF, 0,   3, INF}, 
                          {INF, INF, 0,   1}, 
                          {INF, INF, INF, 0} 
                        }; 
        AllPairShortestPath a = new AllPairShortestPath(); 
  
        // Print the solution 
        a.floydWarshall(graph); 
    } 
} 
  
// Contributed by Aakash Hasija 

Python

# Python Program for Floyd Warshall Algorithm 
  
# Number of vertices in the graph 
V = 4 
  
# Define infinity as the large enough value. This value will be 
# used for vertices not connected to each other 
INF  = 99999
  
# Solves all pair shortest path via Floyd Warshall Algorithm 
def floydWarshall(graph): 
  
    """ dist[][] will be the output matrix that will finally 
        have the shortest distances between every pair of vertices """
    """ initializing the solution matrix same as input graph matrix 
    OR we can say that the initial values of shortest distances 
    are based on shortest paths considering no  
    intermediate vertices """
    dist = map(lambda i : map(lambda j : j , i) , graph) 
      
    """ 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 the set  
    {0, 1, 2, .. k-1} as intermediate vertices. 
      ----> After the end of a iteration, vertex no. k is 
     added to the set of intermediate vertices and the  
    set becomes {0, 1, 2, .. k} 
    """
    for k in range(V): 
  
        # pick all vertices as source one by one 
        for i in range(V): 
  
            # Pick all vertices as destination for the 
            # above picked source 
            for j in range(V): 
  
                # If vertex k is on the shortest path from  
                # i to j, then update the value of dist[i][j] 
                dist[i][j] = min(dist[i][j] , 
                                  dist[i][k]+ dist[k][j] 
                                ) 
    printSolution(dist) 
  
  
# A utility function to print the solution 
def printSolution(dist): 
    print "Following matrix shows the shortest distances\ 
 between every pair of vertices" 
    for i in range(V): 
        for j in range(V): 
            if(dist[i][j] == INF): 
                print "%7s" %("INF"), 
            else: 
                print "%7d\t" %(dist[i][j]), 
            if j == V-1: 
                print "" 
  
  
  
# Driver program to test the above program 
# Let us create the following weighted graph 
""" 
            10 
       (0)------->(3) 
        |         /|\ 
      5 |          | 
        |          | 1 
       \|/         | 
       (1)------->(2) 
            3           """
graph = [[0,5,INF,10], 
             [INF,0,3,INF], 
             [INF, INF, 0,   1], 
             [INF, INF, INF, 0] 
        ] 
# Print the solution 
floydWarshall(graph); 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

C#

// A C# program for Floyd Warshall All  
// Pairs Shortest Path algorithm. 
  
using System; 
  
public class AllPairShortestPath 
{ 
    readonly static int INF = 99999, V = 4; 
  
    void floydWarshall(int[,] graph) 
    { 
        int[,] dist = new int[V, V]; 
        int 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 a 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 a 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); 
    } 
  
    void printSolution(int[,] dist) 
    { 
        Console.WriteLine("Following matrix shows the shortest "+ 
                        "distances between every pair of vertices"); 
        for (int i = 0; i < V; ++i) 
        { 
            for (int j = 0; j < V; ++j) 
            { 
                if (dist[i, j] == INF) { 
                    Console.Write("INF "); 
                } else { 
                    Console.Write(dist[i, j] + " "); 
                } 
            } 
              
            Console.WriteLine(); 
        } 
    } 
  
    // Driver Code 
    public static void Main(string[] args) 
    { 
        /* Let us create the following 
           weighted graph 
              10 
        (0)------->(3) 
        |         /|\ 
        5 |         | 
        |         | 1 
        \|/         | 
        (1)------->(2) 
             3             */
        int[,] graph = { {0, 5, INF, 10}, 
                        {INF, 0, 3, INF}, 
                        {INF, INF, 0, 1}, 
                        {INF, INF, INF, 0} 
                        }; 
          
        AllPairShortestPath a = new AllPairShortestPath(); 
  
        // Print the solution 
        a.floydWarshall(graph); 
    } 
} 
  
// This article is contributed by  
// Abdul Mateen Mohammed 

Output:

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

Time Complexity: O(V^3)

The above program only prints the shortest distances. We can modify the solution to print the shortest paths also by storing the predecessor information in a separate 2D matrix.
Also, the value of INF can be taken as INT_MAX from limits.h to make sure that we handle maximum possible value. When we take INF as INT_MAX, we need to change the if condition in the above program to avoid arithmetic overflow.

#include 

#define INF INT_MAX
..........................
if ( dist[i][k] != INF && 
     dist[k][j] != INF && 
     dist[i][k] + dist[k][j] < dist[i][j]
    )
 dist[i][j] = dist[i][k] + dist[k][j];
...........................

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