Floyd Warshall Algorithm | DP-16

• Difficulty Level : Medium
• Last Updated : 12 Jan, 2022

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 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.
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] 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. Following is implementations of the Floyd Warshall algorithm.

C++

 // C++ Program for Floyd Warshall Algorithm #include using namespace std;   // 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 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 code 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; }   // This code is contributed by Mythri J L

C

 // C Program for Floyd Warshall Algorithm #include   // 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 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 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 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(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

Python3

 # 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 Algorithm   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 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 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"),end=" ")             else:                 print ("%7d\t" % (dist[i][j]),end=' ')             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 Mythri J L

C#

 // A 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 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();           // Print the solution         a.floydWarshall(graph);     } }   // This article is contributed by // Abdul Mateen Mohammed

PHP

 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";     } }   // Driver Code   // Number of vertices in the graph \$V = 4 ;   /* Define Infinite as a large enough value. This value will be used for vertices 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));   // Print the solution floydWarshall(\$graph, \$V, \$INF);   // This code is contributed by Ryuga ?>

Javascript



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 arithmetic overflow.

#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];
...........................