Detect Cycle in a an Undirected Graph

Data StructureGraph AlgorithmsAlgorithms

To detect if there is any cycle in the undirected graph or not, we will use the DFS traversal for the given graph. For every visited vertex v, when we have found any adjacent vertex u, such that u is already visited, and u is not the parent of vertex v. Then one cycle is detected. 

We will assume that there are no parallel edges for any pair of vertices.

Input and Output:
Adjacency matrix
   
    0 1 0 0 0
    1 0 1 1 0
    0 1 0 0 1
    0 1 0 0 1
    0 0 1 1 0

Output:
The graph has cycle.

Algorithm

dfs(vertex, visited, parent)

Input: The start vertex, the visited set, and the parent node of the vertex.

Output: True a cycle is found.Begin    add vertex in the visited set    for all vertex v which is adjacent with vertex, do       if v = parent, then          return true       if v is not in the visited set, then          return true       if dfs(v, visited, vertex) is true, then          return true    done    return false End hasCycle(graph) Input: The given graph. Output: True when a cycle has found.Begin    for all vertex v in the graph, do       if v is not in the visited set, then          go for next iteration       if dfs(v, visited, φ) is true, then     //parent of v is null          return true       return false    done End

Example

#include<iostream>
#include<set>
#define NODE 5
using namespace std;

int graph[NODE][NODE] = {
   {0, 1, 0, 0, 0},
   {1, 0, 1, 1, 0},
   {0, 1, 0, 0, 1},
   {0, 1, 0, 0, 1},
   {0, 0, 1, 1, 0}
};

bool dfs(int vertex, set<int>&visited, int parent) {
   visited.insert(vertex);
   for(int v = 0; v<NODE; v++) {
      if(graph[vertex][v]) {
         if(v == parent)    //if v is the parent not move that direction
            continue;
         if(visited.find(v) != visited.end())    //if v is already visited
            return true;
         if(dfs(v, visited, vertex))
            return true;
      }
   }
   return false;
}

bool hasCycle() {
   set<int> visited;       //visited set
   for(int v = 0; v<NODE; v++) {
      if(visited.find(v) != visited.end())    //when visited holds v, jump to next iteration
         continue;
      if(dfs(v, visited, -1)) {    //-1 as no parent of starting vertex
         return true;
      }
   }
   return false;
}

int main() {
   bool res;
   res = hasCycle();
   if(res)
      cout << "The graph has cycle." << endl;
   else
      cout << "The graph has no cycle." << endl;
}

Output

The graph has cycle.
raja
Published on 10-Jul-2018 12:35:28
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