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.

Image Source : http://algo.epfl.ch

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++ Program to check if there is a negative weight // cycle using Floyd Warshall Algorithm #include<iostream> 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]); // 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 verex 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 program int 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; }

Output:

Yes

This article is contributed by **Shivani Mittal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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