# Floyd Warshalls Algorithm

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## What is Floyd Warshalls Algorithm? Image Reference: Geeks for Geeks

Let the vertices of G be V = {1, 2........n} and consider a subset {1, 2........k} of vertices for some k. For any pair of vertices i, j ? V, considered all paths from i to j whose intermediate vertices are all drawn from {1, 2.......k}, and let p be a minimum weight path from amongst them. The Floyd-Warshall algorithm is known to exploits a link between path p and shortest paths from i to j with all intermediate vertices in the set {1, 2.......k-1}. The link depends on whether or not k is an intermediate between vertex of path p.

If k is not the intermediate vertex of path p, then all intermediate vertices of path p are in the set {1, 2........k-1}. Thus, the shortest path from vertex i to vertex j with all intermediate vertices in the set {1, 2.......k-1} is also the shortest path i to j with all intermediate vertices in the set {1, 2.......k}. If k is the intermediate vertex of path p, then we break p down into i ? k ? j.

Let dij(k) be the weight of the shortest path from vertex i to vertex j with all intermediate vertices in the set {1, 2.......k}.

A recursive definition for floyd warshall is given as: ## Floyd Warshalls Flowchart: Image Reference: Geeks for Geeks

## Floyd Warshalls Pseudo Code:

FLOYD - WARSHALL (W) 1. n = rows [W]. 2. D<sup>0</sup> = W 3. for k = 1 to n 4. do for i = 1 to n 5. do for j = 1 to n 6. do dij(k) = min (dij(k-1),dik(k-1)+dkj(k-1) ) 7. return D(n)

## Floyd Warshalls Implementation in Java:

import java.util.*; import java.lang.*; import java.io.*; class AllPairShortestPath { /*final static int */ final static int INF = 99999, V = 4; void floydWarshall(int graph[][]) { int dist[][] = new int[V][V]; int i, j, k; for (i = 0; i < V; i++) for (j = 0; j < V; j++) dist[i][j] = graph[i][j]; for (k = 0; k < V; k++) { for (i = 0; i < V; i++) { for (j = 0; j < V; j++) { if (dist[i][k] + dist[k][j] < dist[i][j]) dist[i][j] = dist[i][k] + dist[k][j]; } } } } 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); } }

## 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