The Floyd Warshall Algorithm is for solving all pairs of shortest-path problems. The problem is to find the shortest distances between every pair of vertices in a given edge-weighted directed Graph.
It is an algorithm for finding the shortest path between all the pairs of vertices in a weighted graph. This algorithm follows the dynamic programming approach to find the shortest path.
A C-function for a N x N graph is given below. The function stores the all pair shortest path in the matrix cost [N][N]. The cost matrix of the given graph is available in cost Mat [N][N].
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
Output: Shortest distance matrix
0 5 8 9
INF 0 3 4
INF INF 0 1
INF INF INF 0
Floyd Warshall Algorithm:
# define N 4
void floydwarshall()
{
int cost [N][N];
int i, j, k;
for(i=0; i<N; i++)
for(j=0; j<N; j++)
cost [i][j]= cost Mat [i] [j];
for(k=0; k<N; k++)
{
for(i=0; i<N; i++)
for(j=0; j<N; j++)
if(cost [i][j]> cost [i] [k] + cost [k][j];
cost [i][j]=cost [i] [k]+'cost [k] [i]:
}
//display the matrix cost [N] [N]
}
- 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 one by one pick all vertices 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]
The following figure shows the above optimal substructure property in the all-pairs shortest path problem.
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
// 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#
// 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(" INF "); } else { document.write(" " + 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
Time Complexity: O(V3)
Auxiliary Space: O(1)
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 the maximum possible value. When we take INF as INT_MAX, we need to change the if condition in the above program to avoid arithmetic overflow.
C++
#include#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];........................... |
Java
// Java codefinal int INF = Integer.MAX_VALUE;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];} |
Python3
if dist[i][k] != numeric_limits.max() and dist[k][j] != numeric_limits.max() and dist[i][k] + dist[k][j] < dist[i][j]: dist[i][j] = dist[i][k] + dist[k][j] |
C#
using System;const int INF = int.MaxValue;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];} |
Javascript
// js codeif (dist[i][k] !== Number.MAX_SAFE_INTEGER && dist[k][j] !== Number.MAX_SAFE_INTEGER && dist[i][k] + dist[k][j] < dist[i][j]) { dist[i][j] = dist[i][k] + dist[k][j];} |
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