Open In App
Related Articles

Floyd Warshall Algorithm | DP-16

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

Recommended Practice

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] [i];
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 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  // 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 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 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

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";    }} // Drivers' Code // Number of vertices in the graph\$V = 4 ; /* Define Infinite as a large enoughvalue. 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?>

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
"          );          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("
");          }        }      }      // 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.

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];}

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above