# C++ programming Articulation Points (or Cut Vertices) in a Graph

C++ programming Articulation Points (or Cut Vertices) in a Graph - Articulation points represent vulnerabilities in a connected network

A vertex in an undirected connected graph is an articulation point (or cut vertex) iff removing it (and edges through it) disconnects the graph. Articulation points represent vulnerabilities in a connected network – single points whose failure would split the network into 2 or more disconnected components. They are useful for designing reliable networks.
For a disconnected undirected graph, an articulation point is a vertex removing which increases number of connected components.

Following are some example graphs with articulation points encircled with red color.

How to find all articulation points in a given graph?
A simple approach is to one by one remove all vertices and see if removal of a vertex causes disconnected graph. Following are steps of simple approach for connected graph.

1) For every vertex v, do following
…..a) Remove v from graph
..…b) See if the graph remains connected (We can either use BFS or DFS)
…..c) Add v back to the graph

Time complexity of above method is O(V*(V+E)) for a graph represented using adjacency list. Can we do better?

A O(V+E) algorithm to find all Articulation Points (APs)
The idea is to use DFS (Depth First Search). In DFS, we follow vertices in tree form called DFS tree. In DFS tree, a vertex u is parent of another vertex v, if v is discovered by u (obviously v is an adjacent of u in graph). In DFS tree, a vertex u is articulation point if one of the following two conditions is true.

1) u is root of DFS tree and it has at least two children.
2) u is not root of DFS tree and it has a child v such that no vertex in subtree rooted with v has a back edge to one of the ancestors (in DFS tree) of u.

READ  Python Programming - Floyd Warshall Algorithm

Following figure shows same points as above with one additional point that a leaf in DFS Tree can never be an articulation point. (Source Ref 2)

We do DFS traversal of given graph with additional code to find out Articulation Points (APs). In DFS traversal, we maintain a parent[] array where parent[u] stores parent of vertex u. Among the above mentioned two cases, the first case is simple to detect. For every vertex, count children. If currently visited vertex u is root (parent[u] is NIL) and has more than two children, print it.

How to handle second case? The second case is trickier. We maintain an array disc[] to store discovery time of vertices. For every node u, we need to find out the earliest visited vertex (the vertex with minimum discovery time) that can be reached from subtree rooted with u. So we maintain an additional array low[] which is defined as follows.

```low[u] = min(disc[u], disc[w])
where w is an ancestor of u and there is a back edge from
some descendant of u to w.```

C++ programming:

``````// A C++ program to find articulation points in an undirected graph
#include<iostream>
#include <list>
#define NIL -1
using namespace std;

// A class that represents an undirected graph
class Graph
{
int V;    // No. of vertices
void APUtil(int v, bool visited[], int disc[], int low[],
int parent[], bool ap[]);
public:
Graph(int V);   // Constructor
void addEdge(int v, int w);   // function to add an edge to graph
void AP();    // prints articulation points
};

Graph::Graph(int V)
{
this->V = V;
}

{
adj[w].push_back(v);  // Note: the graph is undirected
}

// A recursive function that find articulation points using DFS traversal
// 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
// ap[] --> Store articulation points
void Graph::APUtil(int u, bool visited[], int disc[],
int low[], int parent[], bool ap[])
{
// A static variable is used for simplicity, we can avoid use of static
// variable by passing a pointer.
static int time = 0;

// Count of children in DFS Tree
int children = 0;

// Mark the current node as visited
visited[u] = true;

// Initialize discovery time and low value
disc[u] = low[u] = ++time;

// Go through all vertices aadjacent to this
list<int>::iterator i;
{
int v = *i;  // v is current adjacent of u

// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
APUtil(v, visited, disc, low, parent, ap);

// 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] == NIL && children > 1)
ap[u] = 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] != NIL && low[v] >= disc[u])
ap[u] = true;
}

// Update low value of u for parent function calls.
else if (v != parent[u])
low[u]  = min(low[u], disc[v]);
}
}

// The function to do DFS traversal. It uses recursive function APUtil()
void Graph::AP()
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
bool *ap = new bool[V]; // To store articulation points

// Initialize parent and visited, and ap(articulation point) arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
ap[i] = false;
}

// Call the recursive helper function to find articulation points
// in DFS tree rooted with vertex 'i'
for (int i = 0; i < V; i++)
if (visited[i] == false)
APUtil(i, visited, disc, low, parent, ap);

// Now ap[] contains articulation points, print them
for (int i = 0; i < V; i++)
if (ap[i] == true)
cout << i << " ";
}

// Driver program to test above function
int main()
{
// Create graphs given in above diagrams
cout << "\nArticulation points in first graph \n";
Graph g1(5);
g1.AP();

cout << "\nArticulation points in second graph \n";
Graph g2(4);
g2.AP();

cout << "\nArticulation points in third graph \n";
Graph g3(7);
g3.AP();

return 0;
}``````

Output:

```Articulation points in first graph
0 3
Articulation points in second graph
1 2
Articulation points in third graph
1``` #### Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.