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.

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 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); } /* A utility function to print solution */ void printSolution(int dist[][V]) { printf ("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 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[][]) { 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.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 Algrorithm 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 considerting no intermedidate 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 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 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)

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 arithmatic overflow.

#include <limits.h> #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