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 update 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 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 is taken from the Cormen book. It shows the above optimal substructure property in the all-pairs shortest path problem.
package DP;
public class FloydWarshallAlgorithm {
public static final int V = 4; // Number of vertices in the graph
/* Define Infinite as a large enough value. This value will be used
for vertices not connected to each other */
public static final int INF = 99999;
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}
};
floydWarshell(graph);
}
// Solves the all-pairs shortest path problem using Floyd Warshall algorithm
public static void floydWarshell(int[][] graph){
int V = graph[0].length;
/* dist[][] will be the output matrix that will finally have the shortest
distances between every pair of vertices */
int[][] dist = new int[V][V];
/* 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(int i=0; i<V; i++){
for(int 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(int k=0; k<V; k++){
for(int i=0; i<V; i++){ // Pick all vertices as source one by one
for(int j=0; j<V; j++){ // Pick all vertices as destination for the above picked source
// If vertex k is on the shortest path from
// i to j, then update the value of dist[i][j]
dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
}
}
}
// Print the shortest distance matrix
printSolution(dist);
}
/* A utility function to print solution */
public static void printSolution(int[][] dist){
System.out.println("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.format("%7s", "INF");
}else{
System.out.format("%8d", dist[i][j]);
}
}
System.out.println();
}
}
}