Try it on GfG Practice

Given a matrixdist[][] of sizen x n, wheredist[i][j] represents the weight of the edge from nodei to nodej. If there is no direct edge,dist[i][j] is set to a large value (e.g., 10⁸) to represent infinity. The diagonal entriesdist[i][i] are0, since the distance from a node to itself is zero. The graphmay contain negative edge weights, but itdoes not contain any negative weight cycles.

Your task is to determine theshortest path distance between all pair of nodesi andjin the graph.

Example:

Input: dist[][] = [[0, 4, 10⁸, 5, 10⁸],
[10⁸, 0, 1, 10⁸, 6],
[2, 10⁸, 0, 3, 10⁸],
[10⁸, 10⁸, 1, 0, 2],
[1, 10⁸, 10⁸, 4, 0]]

Output:[[0, 4, 5, 5, 7],
[3, 0, 1, 4, 6],
[2, 6, 0, 3, 5],
[3, 7, 1, 0, 2],
[1, 5, 5, 4, 0]]
Explanation:

Each celldist[i][j] in the output shows the shortest distance from nodei to nodej, computed by considering all possible intermediate nodes using the Floyd-Warshall algorithm.

Floyd Warshall Algorithm:

TheFloyd–Warshall algorithm works by maintaining a two-dimensional array that represents the distances between nodes. Initially, this array is filled using only the direct edges between nodes. Then, the algorithm gradually updates these distances by checking if shorter paths exist through intermediate nodes.

This algorithm works for both thedirected andundirected weighted graphs and can handle graphs with bothpositive andnegative weight edges.

Note: It does not work for the graphs with negative cycles (where the sum of the edges in a cycle is negative).

Idea Behind Floyd Warshall Algorithm:

Suppose we have a graph dist[][]withV vertices from 0 toV-1. Now we have to evaluate adist[][]wheredist[i][j]represents the shortest path between vertexi toj.

Let us assume that verticesitoj have intermediate nodes. The idea behind Floyd Warshall algorithm is to treat each and every vertex k from0toV-1 as an intermediate node one by one. When we consider the vertex k, we must have considered vertices from 0 to k-1 already. So we use the shortest paths built by previous vertices to build shorter paths with vertex k included.

The following figure shows the above optimal substructure property in Floyd Warshall algorithm:

Why Floyd Warshall Works (Correctness Proof)?

The algorithm relies on theprinciple of optimal substructure, meaning:

• If the shortest path from i to j passes through some vertex k, then the path from i to k and the path from k to j must also be shortest paths.
• The iterative approach ensures that by the time vertex k is considered, all shortest paths using only vertices 0 to k-1 have already been computed.

By the end of the algorithm, all shortest paths are computed optimally because each possible intermediate vertex has been considered.

Why Floyd-Warshall Algorithm better for Dense Graphs and not for Sparse Graphs?

Dense Graph: A graph in which the number of edges are significantly much higher than the number of vertices.
Sparse Graph: A graph in which the number of edges are very much low.

No matter how many edges are there in the graph theFloyd Warshall Algorithm runs for O(V³) times therefore it is best suited forDense graphs. In the case of sparse graphs,Johnson's Algorithm is more suitable.

Step-by-step implementation

• Start by updating the distance matrix by treating each vertex as a possible intermediate node between all pairs of vertices.
• Iterate through each vertex, one at a time. For each selected vertexk, attempt to improve the shortest paths that pass through it.
• When we pick vertex numberk 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 ito j. We keep the value of dist[i][j]as it is. 
  • k is an intermediate vertex in shortest path fromi to j. We update the value of dist[i][j]asdist[i][k] + dist[k][j], ifdist[i][j] > dist[i][k] + dist[k][j]
• Repeat this process for each vertexk until all intermediate possibilities have been considered.

Illustration:

#include<iostream>#include<vector>#include<climits>usingnamespacestd;// Solves the all-pairs shortest path// problem using Floyd Warshall algorithmvoidfloydWarshall(vector<vector<int>>&dist){intV=dist.size();// Add all vertices one by one to// the set of intermediate vertices.for(intk=0;k<V;k++){// Pick all vertices as source one by onefor(inti=0;i<V;i++){// Pick all vertices as destination// for the above picked sourcefor(intj=0;j<V;j++){// shortest path from// i to jif(dist[i][k]!=1e8&&dist[k][j]!=1e8)dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);}}}}intmain(){intINF=100000000;vector<vector<int>>dist={{0,4,INF,5,INF},{INF,0,1,INF,6},{2,INF,0,3,INF},{INF,INF,1,0,2},{1,INF,INF,4,0}};floydWarshall(dist);for(inti=0;i<dist.size();i++){for(intj=0;j<dist.size();j++){cout<<dist[i][j]<<" ";}cout<<endl;}return0;}

importjava.util.*;classGfG{// Solves the all-pairs shortest path// problem using Floyd Warshall algorithmstaticvoidfloydWarshall(int[][]dist){intV=dist.length;// Add all vertices one by one to// the set of intermediate vertices.for(intk=0;k<V;k++){// Pick all vertices as source one by onefor(inti=0;i<V;i++){// Pick all vertices as destination// for the above picked sourcefor(intj=0;j<V;j++){// shortest path from// i to jif(dist[i][k]!=1e8&&dist[k][j]!=1e8)dist[i][j]=Math.min(dist[i][j],dist[i][k]+dist[k][j]);}}}}publicstaticvoidmain(String[]args){intINF=100000000;int[][]dist={{0,4,INF,5,INF},{INF,0,1,INF,6},{2,INF,0,3,INF},{INF,INF,1,0,2},{1,INF,INF,4,0}};floydWarshall(dist);for(inti=0;i<dist.length;i++){for(intj=0;j<dist.length;j++){System.out.print(dist[i][j]+" ");}System.out.println();}}}

# Solves the all-pairs shortest path# problem using Floyd Warshall algorithmdeffloydWarshall(dist):V=len(dist)# Add all vertices one by one to# the set of intermediate vertices.forkinrange(V):# Pick all vertices as source one by oneforiinrange(V):# Pick all vertices as destination# for the above picked sourceforjinrange(V):#shortest path from#i to jif(dist[i][k]!=100000000anddist[k][j]!=100000000):dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);if__name__=="__main__":INF=100000000;dist=[[0,4,INF,5,INF],[INF,0,1,INF,6],[2,INF,0,3,INF],[INF,INF,1,0,2],[1,INF,INF,4,0]]floydWarshall(dist)foriinrange(len(dist)):forjinrange(len(dist)):print(dist[i][j],end=" ")print()

usingSystem;classGfG{staticvoidfloydWarshall(int[,]dist){intV=dist.GetLength(0);for(intk=0;k<V;k++){for(inti=0;i<V;i++){for(intj=0;j<V;j++){// shortest path from// i to jif(dist[i,k]!=1e8&&dist[k,j]!=1e8)dist[i,j]=Math.Min(dist[i,j],dist[i,k]+dist[k,j]);}}}}// large number as "infinity"constintINF=100000000;staticvoidMain(){int[,]dist={{0,4,INF,5,INF},{INF,0,1,INF,6},{2,INF,0,3,INF},{INF,INF,1,0,2},{1,INF,INF,4,0}};floydWarshall(dist);for(inti=0;i<dist.GetLength(0);i++){for(intj=0;j<dist.GetLength(1);j++){Console.Write(dist[i,j]+" ");}Console.WriteLine();}}}

// Solves the all-pairs shortest path// problem using Floyd Warshall algorithmfunctionfloydWarshall(dist){letV=dist.length;// Add all vertices one by one to// the set of intermediate vertices.for(letk=0;k<V;k++){// Pick all vertices as source one by onefor(leti=0;i<V;i++){// Pick all vertices as destination// for the above picked sourcefor(letj=0;j<V;j++){// shortest path from// i to jif(dist[i][k]!=INF&&dist[k][j]!=INF){dist[i][j]=Math.min(dist[i][j],dist[i][k]+dist[k][j]);}}}}}letINF=100000000;// Driver Codeletdist=[[0,4,INF,5,INF],[INF,0,1,INF,6],[2,INF,0,3,INF],[INF,INF,1,0,2],[1,INF,INF,4,0]];floydWarshall(dist);for(leti=0;i<dist.length;i++){console.log(dist[i].join(" "));}

0 4 5 5 7 3 0 1 4 6 2 6 0 3 5 3 7 1 0 2 1 5 5 4 0

Time Complexity:O(V³), where V is the number of vertices in the graph and we run three nested loops each of size V.
Auxiliary Space:O(1).

Read here for detailed analysis: complexity analysis of the Floyd Warshall algorithm

Note: 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. 

Real World Applications of Floyd-Warshall Algorithm

• In computer networking, the algorithm can be used to find the shortest path between all pairs of nodes in a network. This is termed asnetwork routing.
• Flight Connectivity In the aviation industry to find the shortest path between the airports.
• GIS(Geographic Information Systems) applications often involve analyzing spatial data, such as road networks, to find the shortest paths between locations.
• Kleene's algorithm which is a generalization of floyd warshall, can be used to find regular expression for a regular language.

Important Interview questions related to Floyd-Warshall

• How to Detect Negative Cycle in a graph using Floyd Warshall Algorithm?
• How is Floyd-warshall algorithm different from Dijkstra's algorithm?
• How is Floyd-warshall algorithm different from Bellman-Ford algorithm?

Problems based on Shortest Path

• Shortest Path in Directed Acyclic Graph
• Shortest path with one curved edge in an undirected Graph
• Minimum Cost Path
• Path with smallest difference between consecutive cells
• Print negative weight cycle in a Directed Graph
• 1st to Kth shortest path lengths in given Graph
• Shortest path in a Binary Maze
• Minimum steps to reach target by a Knight
• Number of ways to reach at destination in shortest time
• Snake and Ladder Problem
• Word Ladder