Detecting negative cycle using Floyd Warshall

• Difficulty Level : Easy
• Last Updated : 24 Nov, 2021

We are given a directed graph. We need compute whether the graph has negative cycle or not. A negative cycle is one in which the overall sum of the cycle comes negative. Negative weights are found in various applications of graphs. For example, instead of paying cost for a path, we may get some advantage if we follow the path.
Examples:

Input : 4 4
0 1 1
1 2 -1
2 3 -1
3 0 -1

Output : Yes
The graph contains a negative cycle. We have discussed Bellman Ford Algorithm based solution for this problem.
In this post, Floyd Warshall Algorithm based solution is discussed that works for both connected and disconnected graphs.
Distance of any node from itself is always zero. But in some cases, as in this example, when we traverse further from 4 to 1, the distance comes out to be -2, i.e. distance of 1 from 1 will become -2. This is our catch, we just have to check the nodes distance from itself and if it comes out to be negative, we will detect the required negative cycle.

C++

 // C++ Program to check if there is a negative weight// cycle using Floyd Warshall Algorithm#includeusing 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 matrixvoid printSolution(int dist[][V]);   // Returns true if graph has negative weight cycle// else false.bool negCyclefloydWarshall(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 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];            }        }    }      // If distance of any vertex from itself    // becomes negative, then there is a negative    // weight cycle.    for (int i = 0; i < V; i++)        if (dist[i][i] < 0)            return true;    return false; }    // driver programint main(){    /* Let us create the following weighted graph            1    (0)----------->(1)    /|\             |     |              |  -1 |              | -1     |             \|/    (3)<-----------(2)        -1       */              int graph[V][V] = { {0   , 1   , INF , INF},                        {INF , 0   , -1  , INF},                        {INF , INF , 0   ,  -1},                        {-1  , INF , INF ,   0}};          if (negCyclefloydWarshall(graph))       cout << "Yes";    else       cout << "No";     return 0;}

Java

 // Java Program to check if there is a negative weight// cycle using Floyd Warshall Algorithmclass GFG{      // Number of vertices in the graph    static final int V = 4;          /* Define Infinite as a large enough value. This     value will be used for vertices not connected     to each other */    static final int INF = 99999;          // Returns true if graph has negative weight cycle    // else false.    static boolean negCyclefloydWarshall(int graph[][])    {                  /* dist[][] will be the output matrix that will         finally have the shortest         distances between every pair of vertices */        int dist[][] = new int[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 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];                }            }        }              // If distance of any vertex from itself        // becomes negative, then there is a negative        // weight cycle.        for (i = 0; i < V; i++)            if (dist[i][i] < 0)                return true;          return false;     }              // Driver code    public static void main (String[] args)    {          /* Let us create the following weighted graph                1        (0)----------->(1)        /|\               |         |               |      -1 |               | -1         |                \|/        (3)<-----------(2)            -1     */                      int graph[][] = { {0, 1, INF, INF},                          {INF, 0, -1, INF},                          {INF, INF, 0, -1},                          {-1, INF, INF, 0}};                  if (negCyclefloydWarshall(graph))            System.out.print("Yes");        else            System.out.print("No");     }}  // This code is contributed by Anant Agarwal.

Python3

 # Python Program to check# if there is a# negative weight# cycle using Floyd# Warshall Algorithm     # Number of vertices# in the graphV = 4       # Define Infinite as a# large enough value. This # value will be used #for vertices not connected # to each other INF = 99999       # Returns true if graph has# negative weight cycle# else false.def negCyclefloydWarshall(graph):               # dist[][] will be the    # output matrix that will     # finally have the shortest     # distances between every    # pair of vertices     dist=[[0 for i in range(V+1)]for j in range(V+1)]           # 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 in range(V):        for j in range(V):            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 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]                if (dist[i][k] + dist[k][j] < dist[i][j]):                        dist[i][j] = dist[i][k] + dist[k][j]       # If distance of any    # vertex from itself    # becomes negative, then    # there is a negative    # weight cycle.    for i in range(V):        if (dist[i][i] < 0):            return True       return False              # Driver code         ''' Let us create the    following weighted graph            1    (0)----------->(1)    /|\               |     |               |  -1 |               | -1     |                \|/    (3)<-----------(2)        -1     '''           graph = [ [0, 1, INF, INF],          [INF, 0, -1, INF],          [INF, INF, 0, -1],          [-1, INF, INF, 0]]           if (negCyclefloydWarshall(graph)):    print("Yes")else:    print("No")   # This code is contributed# by Anant Agarwal.

C#

 // C# Program to check if there// is a negative weight cycle// using Floyd Warshall Algorithm  using System;  namespace Cycle{public class GFG{                 // Number of vertices in the graph    static int V = 4;          /* Define Infinite as a large enough value. This     value will be used for vertices not connected     to each other */    static int INF = 99999;          // Returns true if graph has negative weight cycle    // else false.    static bool negCyclefloydWarshall(int [,]graph)    {                  /* dist[][] will be the output matrix that will         finally have the shortest         distances between every pair of vertices */        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];                }            }        }              // If distance of any vertex from itself        // becomes negative, then there is a negative        // weight cycle.        for (i = 0; i < V; i++)            if (dist[i,i] < 0)                return true;          return false;     }              // Driver code    public static void Main()    {          /* Let us create the following weighted graph                1        (0)----------->(1)        /|\             |        |             |    -1 |             | -1        |             \|/        (3)<-----------(2)            -1     */                      int [,]graph = { {0, 1, INF, INF},                        {INF, 0, -1, INF},                        {INF, INF, 0, -1},                        {-1, INF, INF, 0}};                  if (negCyclefloydWarshall(graph))            Console.Write("Yes");        else            Console.Write("No");     }} }  // This code is contributed by Sam007.

Javascript



Output:

Yes

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