Detect a negative cycle in a Graph using the Shortest Path Faster Algorithm

The Shortest Path Faster Algorithm is an improved or more optimized version of the Bellman-Ford Algorithm. It calculates the single source's shortest path in a weighted directed graph. This algorithm is especially suitable for graphs with negatively weighted edges.

Algorithm

Given a weighted directed graph and a source vertex , the algorithm finds the shortest path from , to each vertex , in the graph. The length of thh shortest path from to is stored in for each vertex .

procedure Shortest-Path-Faster-Algorithm(G, s)
   for each vertex v ≠ s in V(G)
      d(v) := ∞
   d(s) := 0
   push s into Q
   while Q is not empty do
      u := poll Q
      for each edge (u, v) in E(G) do
         if d(u) + w(u, v) < d(v) then
            d(v) := d(u) + w(u, v)
            if v is not in Q then
               push v into Q

Problem Statement

Given a graph G containing N nodes having values from 0 to N-1, a source S and an array A[][3] of the type {u, v, w} denoting a directed edge from u to v having weight w. The aim is to find if there exists a negative cycle in the graph from the given source.

Sample Example 1

Input: N = 3, S = 0, A[][] = {{0, 1, -2}, {1, 2, 1}, {2, 0, -1}}
Output: True

Explanation

Starting from 0, the graph contains the following cycle −

0 -> 1 -> 2 -> 0

Sum of weights of the above cycle = (-2) + 1 + (-1) = (-2)

Thus, graph contains cycle of negative weight.

Sample Example 2

Input: N = 3, S = 0, A[][] = {{0, 1, -2}, {1, 2, 1}, {0, 2, -1}}
Output: False

Explanation

Starting from 0, the graph does not contain a cycle.

Solution Approach

Using SPFA to detect a negative weighted cycle in a graph, we follow the following steps −

• Create arrays distance[] with infinity value, visited[] with false and count[] with 0 which will contain the number of times a node has been relaxed.

• Then traverse the graph using the SPFA algorithm.

• Increment the count for each vertex upon relaxation i.e. updating the cost of each vertex connected to v if the cost improves by including v in the path.

• Return true if some vertex is relaxed for the Nth time else returns false.

Example: C++ Implementation

The following code uses the SPFA algorithm to find the negative cycle in the graph.

#include <bits/stdc++.h>
using namespace std;

bool SFPA(int V, int S, int E[][3], int M){
   // Adjacency list of the given graph
   vector<pair<int, int>> g[V];
   // Create Adjacency List
   for (int i = 0; i < M; i++){
      int u = E[i][0];
      int v = E[i][1];
      int w = E[i][2];
      g[u].push_back({v, w});
   }
   vector<int> distance(V, INT_MAX);
   vector<bool> visted(V, false);
   vector<int> count(V, 0);
   // Distance from src to src is 0
   distance[S] = 0;
   queue<int> q;
   q.push(S);
   // Mark source as visited
   visted[S] = true;
   while (!q.empty()) {
      int u = q.front();
      q.pop();
      visted[u] = false;
      // Relaxing all edges of vertex from the Queue
      for (pair<int, int> x : g[u]) {
         int v = x.first;
         int cost = x.second;
         // Update the distance[v] to minimum distance
         if (distance[v] > distance[u] + cost) {
            distance[v] = distance[u] + cost;
            // If vertex v is in Queue
            if (!visted[v]) {
               q.push(v);
               visted[v] = true;
               count[v]++;
               // Negative cycle
               if (count[v] >= V)
               return true;
            }
         }
      }
   }
   // No cycle found
   return false;
}
int main(){
   int N = 4;
   int S = 0;
   int M = 4;
   // Given Edges with weight
   int E[][3] = {{0, 1, 1},
   {1, 2, -1},
   {2, 3, -1},
   {3, 0, -1}};
   // If cycle is present
   if (SFPA(N, S, E, M) == true)
   cout << "True" << endl;
   else
   cout << "False" << endl;
   return 0;
}

Output

True

Conclusion

In conclusion, in order to detect negative cycle in a graph, Shortest Path Fast Algorithm can be used as it is most efficient when dealing with negative weights of edges. The above solution gives a time complexity of O(N*M) and a space complexity of O(N).