PYTHON Programming - Biconnected graph - An undirected graph is called Biconnected if there are two vertex - disjoint paths between any two vertices.

An undirected graph is called Biconnected if there are two vertex-disjoint paths between any two vertices. In a Biconnected Graph, there is a simple cycle through any two vertices.

By convention, two nodes connected by an edge form a biconnected graph, but this does not verify the above properties. For a graph with more than two vertices, the above properties must be there for it to be Biconnected.

Following are some examples.

**How to find if a given graph is Biconnected or not?**

*A connected graph is Biconnected if it is connected and doesn’t have any Articulation Point*. We mainly need to check two things in a graph.

1) The graph is connected.

2) There is not articulation point in graph.

We start from any vertex and do DFS traversal. In DFS traversal, we check if there is any articulation point. If we don’t find any articulation point, then the graph is Biconnected. Finally, we need to check whether all vertices were reachable in DFS or not. If all vertices were not reachable, then the graph is not even connected.

```
# Python program to find if a given undirected graph is
# biconnected
from collections import defaultdict
#This class represents an 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
self.Time = 0
# function to add an edge to graph
def addEdge(self,u,v):
self.graph[u].append(v)
self.graph[v].append(u)
'''A recursive function that returns true if there is an articulation
point in given graph, otherwise returns false.
This function is almost same as isAPUtil()
u --> The vertex to be visited next
visited[] --> keeps tract of visited vertices
disc[] --> Stores discovery times of visited vertices
parent[] --> Stores parent vertices in DFS tree'''
def isBCUtil(self,u, visited, parent, low, disc):
#Count of children in current node
children =0
# Mark the current node as visited and print it
visited[u]= True
# Initialize discovery time and low value
disc[u] = self.Time
low[u] = self.Time
self.Time += 1
#Recur for all the vertices adjacent to this vertex
for v in self.graph[u]:
# If v is not visited yet, then make it a child of u
# in DFS tree and recur for it
if visited[v] == False :
parent[v] = u
children += 1
if self.isBCUtil(v, visited, parent, low, disc):
return True
# Check if the subtree rooted with v has a connection to
# one of the ancestors of u
low[u] = min(low[u], low[v])
# u is an articulation point in following cases
# (1) u is root of DFS tree and has two or more chilren.
if parent[u] == -1 and children > 1:
return True
#(2) If u is not root and low value of one of its child is more
# than discovery value of u.
if parent[u] != -1 and low[v] >= disc[u]:
return True
elif v != parent[u]: # Update low value of u for parent function calls.
low[u] = min(low[u], disc[v])
return False
# The main function that returns true if graph is Biconnected,
# otherwise false. It uses recursive function isBCUtil()
def isBC(self):
# Mark all the vertices as not visited and Initialize parent and visited,
# and ap(articulation point) arrays
visited = [False] * (self.V)
disc = [float("Inf")] * (self.V)
low = [float("Inf")] * (self.V)
parent = [-1] * (self.V)
# Call the recursive helper function to find if there is an
# articulation points in given graph. We do DFS traversal starting
# from vertex 0
if self.isBCUtil(0, visited, parent, low, disc):
return False
'''Now check whether the given graph is connected or not.
An undirected graph is connected if all vertices are
reachable from any starting point (we have taken 0 as
starting point)'''
if any(i == False for i in visited):
return False
return True
# Create a graph given in the above diagram
g1 = Graph(2)
g1.addEdge(0, 1)
print "Yes" if g1.isBC() else "No"
g2 = Graph(5)
g2.addEdge(1, 0)
g2.addEdge(0, 2)
g2.addEdge(2, 1)
g2.addEdge(0, 3)
g2.addEdge(3, 4)
g2.addEdge(2, 4)
print "Yes" if g2.isBC() else "No"
g3 = Graph(3)
g3.addEdge(0, 1)
g3.addEdge(1, 2)
print "Yes" if g3.isBC() else "No"
g4 = Graph (5)
g4.addEdge(1, 0)
g4.addEdge(0, 2)
g4.addEdge(2, 1)
g4.addEdge(0, 3)
g4.addEdge(3, 4)
print "Yes" if g4.isBC() else "No"
g5 = Graph(3)
g5.addEdge(0, 1)
g5.addEdge(1, 2)
g5.addEdge(2, 0)
print "Yes" if g5.isBC() else "No"
```

Output:

Yes
Yes
No
No
Yes

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