The Floyd Warshall Algorithm is for solving all pairs shortest path problems. The problem is to find the 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
Output:
Shortest distance matrix
0 5 8 9
INF 0 3 4
INF INF 0 1
INF INF INF 0
- 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 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.
- 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]
// 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 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][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()
{
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 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'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);
}
}
OUTPUT:
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
// 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'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;
}
OUTPUT:
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
# 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" % (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)
OUTPUT:
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
// 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);
}
}
OUTPUT:
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