Floyd Warshall Algorithm - GeeksforGeeks (2024)

Last Updated : 04 Apr, 2024

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The Floyd-Warshall algorithm, named after its creators Robert Floyd and Stephen Warshall, is a fundamental algorithm in computer science and graph theory. It is used to find the shortest paths between all pairs of nodes in a weighted graph. This algorithm is highly efficient and can handle graphs with both positive and negative edge weights, making it a versatile tool for solving a wide range of network and connectivity problems.

Table of Content

Floyd Warshall Algorithm
Idea Behind Floyd Warshall Algorithm
Floyd Warshall Algorithm Algorithm
Pseudo-Code of Floyd Warshall Algorithm
Illustration of Floyd Warshall Algorithm
Complexity Analysis of Floyd Warshall Algorithm
Why Floyd-Warshall Algorithm better for Dense Graphs and not for Sparse Graphs?
Important Interview questions related to Floyd-Warshall
Real World Applications of Floyd-Warshall Algorithm

Floyd Warshall Algorithm:

The Floyd Warshall Algorithm is an all pair shortest path algorithm unlike Dijkstra and Bellman Ford which are single source shortest path algorithms. This algorithm works for both the directed and undirected weighted graphs. But, it does not work for the graphs with negative cycles (where the sum of the edges in a cycle is negative). It follows Dynamic Programming approach to check every possible path going via every possible node in order to calculate shortest distance between every pair of nodes.

Idea Behind Floyd Warshall Algorithm:

Suppose we have a graph G[][] with V vertices from 1 to N. Now we have to evaluate a shortestPathMatrix[][] where shortestPathMatrix[i][j] represents the shortest path between vertices i and j.
Obviously the shortest path between i to j will have some k number of intermediate nodes. The idea behind floyd warshall algorithm is to treat each and every vertex from 1 to N as an intermediate node one by one.
The following figure shows the above optimal substructure property in floyd warshall algorithm:

Floyd Warshall Algorithm Algorithm:

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 pick all vertices one by one 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]

Pseudo-Code of Floyd Warshall Algorithm :

For k = 0 to n – 1
For i = 0 to n – 1
For j = 0 to n – 1
Distance[i, j] = min(Distance[i, j], Distance[i, k] + Distance[k, j])
where i = source Node, j = Destination Node, k = Intermediate Node

Illustration of Floyd Warshall Algorithm :

Suppose we have a graph as shown in the image:
See Also
Floyd-Warshall Algorithmus - Kürzeste Wege: Beispiel Floyd-Warshall Algorithm Floyd-Warshall Algorithm | Brilliant Math & Science Wiki Floyd-Warshall - finding all shortest paths
Step 1: Initialize the Distance[][] matrix using the input graph such that Distance[i][j]= weight of edge from i to j, also Distance[i][j] = Infinity if there is no edge from i to j.
Step 2: Treat node A as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:
= Distance[i][j] = minimum (Distance[i][j], (Distance from i to A) + (Distance from A to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][A] + Distance[A][j])
Step 3: Treat node B as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:
= Distance[i][j] = minimum (Distance[i][j], (Distance from i to B) + (Distance from B to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][B] + Distance[B][j])
Step 4: Treat node C as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:
= Distance[i][j] = minimum (Distance[i][j], (Distance from i to C) + (Distance from C to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][C] + Distance[C][j])
Step 5: Treat node D as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:
= Distance[i][j] = minimum (Distance[i][j], (Distance from i to D) + (Distance from D to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][D] + Distance[D][j])
Step 6: Treat node E as an intermediate node and calculate the Distance[][] for every {i,j} node pair using the formula:
= Distance[i][j] = minimum (Distance[i][j], (Distance from i to E) + (Distance from E to j ))
= Distance[i][j] = minimum (Distance[i][j], Distance[i][E] + Distance[E][j])
Step 7: Since all the nodes have been treated as an intermediate node, we can now return the updated Distance[][] matrix as our answer matrix.

Recommended Practice

Floyd Warshall

Try It!

See Also

Floyd Warshall Algorithm

Below is the implementation of the above approach:

C++

// 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 enoughvalue.This value will be used forvertices not connected to each other */#define INF 99999// A function to print the solution matrixvoid printSolution(int dist[][V]);// Solves the all-pairs shortest path// problem using Floyd Warshall algorithmvoid floydWarshall(int dist[][V]){ int i, j, k; /* 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 codeint 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;}// This code is contributed by Mythri J L

// 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 matrixvoid printSolution(int dist[][V]);// Solves the all-pairs shortest path// problem using Floyd Warshall algorithmvoid floydWarshall(int dist[][V]){ int i, j, k; /* 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 codeint 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;}

Java

// 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 dist[][]) { int i, j, k; /* 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); }}// Contributed by Aakash Hasija

Python3

# Python3 Program for Floyd Warshall Algorithm# Number of vertices in the graphV = 4# Define infinity as the large# enough value. This value will be# used for vertices not connected to each otherINF = 99999# Solves all pair shortest path# via Floyd Warshall Algorithmdef 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 solutiondef 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\t" % (dist[i][j]), end=' ') if j == V-1: print()# Driver's codeif __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)# This code is contributed by Mythri J L

// 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); }}// This article is contributed by// Abdul Mateen Mohammed

Javascript

// A JavaScript program for Floyd Warshall All // Pairs Shortest Path algorithm. var INF = 99999; class AllPairShortestPath { constructor() { this.V = 4; } floydWarshall(graph) { var dist = Array.from(Array(this.V), () => new Array(this.V).fill(0)); var 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 < this.V; i++) { for (j = 0; j < this.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 < this.V; k++) { // Pick all vertices as source // one by one for (i = 0; i < this.V; i++) { // Pick all vertices as destination // for the above picked source for (j = 0; j < this.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 this.printSolution(dist); } printSolution(dist) { document.write( "Following matrix shows the shortest " + "distances between every pair of vertices<br>" ); for (var i = 0; i < this.V; ++i) { for (var j = 0; j < this.V; ++j) { if (dist[i][j] == INF) { document.write("&emsp;INF "); } else { document.write("&emsp;&emsp;" + dist[i][j] + " "); } } document.write("<br>"); } } } // Driver Code /* Let us create the following weighted graph 10 (0)------->(3) | /|\ 5 | | | | 1 \|/ | (1)------->(2) 3 */ var graph = [ [0, 5, INF, 10], [INF, 0, 3, INF], [INF, INF, 0, 1], [INF, INF, INF, 0], ]; var a = new AllPairShortestPath(); // Print the solution a.floydWarshall(graph);  // This code is contributed by rdtaank.

PHP

<?php// PHP Program for Floyd Warshall Algorithm // Solves the all-pairs shortest path problem// using Floyd Warshall algorithm function floydWarshall ($graph, $V, $INF) { /* dist[][] will be the output matrix  that will finally have the shortest  distances between every pair of vertices */ $dist = array(array(0,0,0,0), array(0,0,0,0), array(0,0,0,0), array(0,0,0,0)); /* Initialize the solution matrix same  as input graph matrix. Or we can say the  initial values of shortest distances are  based on shortest paths considering no  intermediate vertex. */ for ($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, $V, $INF); } /* A utility function to print solution */function printSolution($dist, $V, $INF) { echo "The following matrix shows the " . "shortest distances between " . "every pair of vertices \n"; for ($i = 0; $i < $V; $i++) { for ($j = 0; $j < $V; $j++) { if ($dist[$i][$j] == $INF) echo "INF " ; else echo $dist[$i][$j], " "; } echo "\n"; } } // Drivers' Code// Number of vertices in the graph $V = 4 ;/* Define Infinite as a large enough value. This value will be used forvertices not connected to each other */$INF = 99999 ;/* Let us create the following weighted graph  10 (0)------->(3)  | /|\ 5 | |  | | 1 \|/ | (1)------->(2)  3 */$graph = array(array(0, 5, $INF, 10), array($INF, 0, 3, $INF), array($INF, $INF, 0, 1), array($INF, $INF, $INF, 0)); // Function callfloydWarshall($graph, $V, $INF); // This code is contributed by Ryuga?>

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

Complexity Analysis of Floyd Warshall Algorithm:

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(V²), to create a 2-D matrix in order to store the shortest distance for each pair of nodes.

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.

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 the Floyd Warshall Algorithm runs for O(V³) times therefore it is best suited for Dense graphs. In the case of sparse graphs, Johnson’s Algorithm is more suitable.

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?

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 as network 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.

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