Detect cycle in an undirected graph-Graph cycle-The time complexity of the union-find algorithm is O(ELogV). Like directed graphs.

Given an undirected graph, how to check if there is a cycle in the graph? For example, the following graph has a cycle 1-0-2-1.

We have discussed cycle detection for directed graph. We have also discussed a union-find algorithm for cycle detection in undirected graphs. The time complexity of the union-find algorithm is O(ELogV). Like directed graphs, we can use DFS to detect cycle in an undirected graph in O(V+E) time. We do a DFS traversal of the given graph. For every visited vertex ‘v’, if there is an adjacent ‘u’ such that u is already visited and u is not parent of v, then there is a cycle in graph. If we don’t find such an adjacent for any vertex, we say that there is no cycle. The assumption of this approach is that there are no parallel edges between any two Vertices

Python Programming:

```
# Python Program to detect cycle in an undirected graph
from collections import defaultdict
#This class represents a undirected graph using adjacency list representation
class Graph:
def __init__(self,vertices):
self.V= vertices #No. of vertices
self.graph = defaultdict(list) # default dictionary to store graph
# function to add an edge to graph
def addEdge(self,v,w):
self.graph[v].append(w) #Add w to v_s list
self.graph[w].append(v) #Add v to w_s list
# A recursive function that uses visited[] and parent to detect
# cycle in subgraph reachable from vertex v.
def isCyclicUtil(self,v,visited,parent):
#Mark the current node as visited
visited[v]= True
#Recur for all the vertices adjacent to this vertex
for i in self.graph[v]:
# If the node is not visited then recurse on it
if visited[i]==False :
if(self.isCyclicUtil(i,visited,v)):
return True
# If an adjacent vertex is visited and not parent of current vertex,
# then there is a cycle
elif parent!=i:
return True
return False
#Returns true if the graph contains a cycle, else false.
def isCyclic(self):
# Mark all the vertices as not visited
visited =[False]*(self.V)
# Call the recursive helper function to detect cycle in different
#DFS trees
for i in range(self.V):
if visited[i] ==False: #Don't recur for u if it is already visited
if(self.isCyclicUtil(i,visited,-1))== True:
return True
return False
# Create a graph given in the above diagram
g = Graph(5)
g.addEdge(1, 0)
g.addEdge(0, 2)
g.addEdge(2, 0)
g.addEdge(0, 3)
g.addEdge(3, 4)
if g.isCyclic():
print "Graph contains cycle"
else :
print "Graph does not contain cycle "
g1 = Graph(3)
g1.addEdge(0,1)
g1.addEdge(1,2)
if g1.isCyclic():
print "Graph contains cycle"
else :
print "Graph does not contain cycle "
```

Output:

Graph contains cycle
Graph doesn't contain cycle

**Time Complexity:** The program does a simple DFS Traversal of graph and graph is represented using adjacency list. So the time complexity is O(V+E)