# Detecting negative cycle using Floyd Warshall

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.

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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.

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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