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
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].
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];
...........................
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Improved By : Abdul Mateen Mohammed
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