Floyd Warshall Algorithm

The Floyd-Warshall algorithm, named after its creators Robert Floyd and Stephen Warshall, is a fundamental algorithm in computer science and graph theory. It is used to find the shortest paths between all pairs of nodes in a weighted graph. This algorithm is highly efficient and can handle graphs with both positive and negative edge weights, making it a versatile tool for solving a wide range of network and connectivity problems.

Floyd Warshall Algorithm:

The Floyd Warshall Algorithm is an all pair shortest path algorithm unlike Dijkstra and Bellman Ford which are single source shortest path algorithms. 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). It follows Dynamic Programming approach to check every possible path going via every possible node in order to calculate shortest distance between every pair of nodes.

Idea Behind Floyd Warshall Algorithm:

Suppose we have a graph G[][] with V vertices from 1 to N. Now we have to evaluate a shortestPathMatrix[][] where shortestPathMatrix[i][j] represents the shortest path between vertices i and j.

Obviously the shortest path between i to j will have some k number of intermediate nodes. The idea behind floyd warshall algorithm is to treat each and every vertex from 1 to N as an intermediate node one by one.

The following figure shows the above optimal substructure property in floyd warshall algorithm:

Floyd Warshall Algorithm Algorithm:

• Initialize the solution matrix same as the input graph matrix as a first step. 
• Then update the solution matrix by considering all vertices as an intermediate vertex. 
• The idea is to pick all vertices one by one 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. 
  • k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is. 
  • 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]

Pseudo-Code of Floyd Warshall Algorithm :

For k = 0 to n – 1
For i = 0 to n – 1
For j = 0 to n – 1
Distance[i, j] = min(Distance[i, j], Distance[i, k] + Distance[k, j])

where i = source Node, j = Destination Node, k = Intermediate Node

Illustration of Floyd Warshall Algorithm :

Suppose we have a graph as shown in the image:

Step 1: Initialize the Distance[][] matrix using the input graph such that Distance[i][j]= weight of edge from i to j, also Distance[i][j] = Infinity if there is no edge from i to j.

Step 2: Treat node A as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:

= Distance[i][j] = minimum (Distance[i][j], (Distance from i to A) + (Distance from A to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][A] + Distance[A][j])

Step 3: Treat node B as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:

= Distance[i][j] = minimum (Distance[i][j], (Distance from i to B) + (Distance from B to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][B] + Distance[B][j])

Step 4: Treat node C as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:

= Distance[i][j] = minimum (Distance[i][j], (Distance from i to C) + (Distance from C to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][C] + Distance[C][j])

Step 5: Treat node D as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:

= Distance[i][j] = minimum (Distance[i][j], (Distance from i to D) + (Distance from D to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][D] + Distance[D][j])

Step 6: Treat node E as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:

= Distance[i][j] = minimum (Distance[i][j], (Distance from i to E) + (Distance from E to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][E] + Distance[E][j])

Step 7: Since all the nodes have been treated as an intermediate node, we can now return the updated Distance[][] matrix as our answer matrix.

Below is the implementation of the above approach:

// C++ Program for Floyd Warshall Algorithm
#include <bits/stdc++.h>
using namespace std;

// 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 dist[][V])
{

    int i, j, k;

    /* 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][j] > (dist[i][k] + dist[k][j])
                    && (dist[k][j] != INF
                        && dist[i][k] != INF))
                    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])
{
    cout << "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)
                cout << "INF"
                     << " ";
            else
                cout << dist[i][j] << "   ";
        }
        cout << endl;
    }
}

// Driver's code
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 } };

    // Function call
    floydWarshall(graph);
    return 0;
}

// This code is contributed by Mythri J L

// 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 dist[][V])
{
    int i, j, k;

    /* 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's code
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 } };

    // Function call
    floydWarshall(graph);
    return 0;
}

// Java program for Floyd Warshall All Pairs Shortest
// Path algorithm.
import java.io.*;
import java.lang.*;
import java.util.*;

class AllPairShortestPath {
    final static int INF = 99999, V = 4;

    void floydWarshall(int dist[][])
    {

        int i, j, k;

        /* 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's 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();

        // Function call
        a.floydWarshall(graph);
    }
}

// Contributed by Aakash Hasija

# Python3 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 = list(map(lambda i: list(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"), end=" ")
            else:
                print("%7d\t" % (dist[i][j]), end=' ')
            if j == V-1:
                print()


# Driver's code
if __name__ == "__main__":
    """
                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]
             ]
    # Function call
    floydWarshall(graph)
# This code is contributed by Mythri J L

// 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's 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();

        // Function call
        a.floydWarshall(graph);
    }
}

// This article is contributed by
// Abdul Mateen Mohammed

// A JavaScript program for Floyd Warshall All
      // Pairs Shortest Path algorithm.

      var INF = 99999;
      class AllPairShortestPath {
        constructor() {
          this.V = 4;
        }

        floydWarshall(graph) {
          var dist = Array.from(Array(this.V), () => new Array(this.V).fill(0));
          var 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 < this.V; i++) {
            for (j = 0; j < this.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 < this.V; k++) {
            // Pick all vertices as source
            // one by one
            for (i = 0; i < this.V; i++) {
              // Pick all vertices as destination
              // for the above picked source
              for (j = 0; j < this.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
          this.printSolution(dist);
        }

        printSolution(dist) {
          document.write(
            "Following matrix shows the shortest " +
              "distances between every pair of vertices<br>"
          );
          for (var i = 0; i < this.V; ++i) {
            for (var j = 0; j < this.V; ++j) {
              if (dist[i][j] == INF) {
                document.write("&emsp;INF ");
              } else {
                document.write("&emsp;&emsp;" + dist[i][j] + " ");
              }
            }

            document.write("<br>");
          }
        }
      }
      // Driver Code
      /* Let us create the following
        weighted graph
            10
        (0)------->(3)
        |         /|\
        5 |         |
        |         | 1
        \|/         |
        (1)------->(2)
            3             */
      var graph = [
        [0, 5, INF, 10],
        [INF, 0, 3, INF],
        [INF, INF, 0, 1],
        [INF, INF, INF, 0],
      ];

      var a = new AllPairShortestPath();

      // Print the solution
      a.floydWarshall(graph);
      
      // This code is contributed by rdtaank.

<?php
// PHP Program for Floyd Warshall Algorithm 

// Solves the all-pairs shortest path problem
// using Floyd Warshall algorithm 
function floydWarshall ($graph, $V, $INF) 
{ 
    /* dist[][] will be the output matrix 
    that will finally have the shortest 
    distances between every pair of vertices */
    $dist = array(array(0,0,0,0),
                  array(0,0,0,0),
                  array(0,0,0,0),
                  array(0,0,0,0));

    /* 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, $V, $INF); 
} 

/* A utility function to print solution */
function printSolution($dist, $V, $INF) 
{ 
    echo "The following matrix shows the " .
             "shortest distances between " . 
                "every pair of vertices \n"; 
    for ($i = 0; $i < $V; $i++) 
    { 
        for ($j = 0; $j < $V; $j++) 
        { 
            if ($dist[$i][$j] == $INF) 
                echo "INF " ; 
            else
                echo $dist[$i][$j], " ";
        } 
        echo "\n"; 
    } 
} 

// Drivers' Code

// Number of vertices in the graph 
$V = 4 ;

/* Define Infinite as a large enough 
value. This value will be used for
vertices not connected to each other */
$INF = 99999 ;

/* Let us create the following weighted graph 
        10 
(0)------->(3) 
    |     /|\ 
5 |     | 
    |     | 1 
\|/     | 
(1)------->(2) 
        3     */
$graph = array(array(0, 5, $INF, 10), 
               array($INF, 0, 3, $INF), 
               array($INF, $INF, 0, 1), 
               array($INF, $INF, $INF, 0)); 

// Function call
floydWarshall($graph, $V, $INF); 

// This code is contributed by Ryuga
?>

The 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   

Complexity Analysis of Floyd Warshall Algorithm:

• Time Complexity: O(V³), where V is the number of vertices in the graph and we run three nested loops each of size V
• Auxiliary Space: O(V²), to create a 2-D matrix in order to store the shortest distance for each pair of nodes.

Note: 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. 

Why Floyd-Warshall Algorithm better for Dense Graphs and not for Sparse Graphs?

Dense Graph: A graph in which the number of edges are significantly much higher than the number of vertices.
Sparse Graph: A graph in which the number of edges are very much low.

No matter how many edges are there in the graph the Floyd Warshall Algorithm runs for O(V³) times therefore it is best suited for Dense graphs. In the case of sparse graphs, Johnson’s Algorithm is more suitable.

Important Interview questions related to Floyd-Warshall:

• How to Detect Negative Cycle in a graph using Floyd Warshall Algorithm?
• How is Floyd-warshall algorithm different from Dijkstra’s algorithm?
• How is Floyd-warshall algorithm different from Bellman-Ford algorithm?

Real World Applications of Floyd-Warshall Algorithm:

• In computer networking, the algorithm can be used to find the shortest path between all pairs of nodes in a network. This is termed as network routing.
• Flight Connectivity In the aviation industry to find the shortest path between the airports.
• GIS(Geographic Information Systems) applications often involve analyzing spatial data, such as road networks, to find the shortest paths between locations.
• Kleene’s algorithm which is a generalization of floyd warshall, can be used to find regular expression for a regular language.
  

GeeksforGeeks

Improve