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Articulation Points (or Cut Vertices) in a Graph

  • Difficulty Level : Hard
  • Last Updated : 19 Jan, 2022

A vertex in an undirected connected graph is an articulation point (or cut vertex) if 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 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. 
 

Articulation Points or Cut Vertices in a GraphArticulation Points or Cut Vertices in a Graph

 

Articulation Points or Cut Vertices in a Graph

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.

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

leaf in DFS Tree can never be an articulation point

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.

Following is the implementation of Tarjan’s algorithm for finding articulation points. 

C++




// C++ program to find articulation points in an undirected graph
#include <bits/stdc++.h>
using namespace std;
 
// A recursive function that find articulation
// points using DFS traversal
// adj[] --> Adjacency List representation of the graph
// u --> The vertex to be visited next
// visited[] --> keeps track of visited vertices
// disc[] --> Stores discovery times of visited vertices
// low[] -- >> earliest visited vertex (the vertex with minimum
// discovery time) that can be reached from subtree
// rooted with current vertex
// parent --> Stores the parent vertex in DFS tree
// isAP[] --> Stores articulation points
void APUtil(vector<int> adj[], int u, bool visited[],
            int disc[], int low[], int& time, int parent,
            bool isAP[])
{
    // 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 adjacent to this
    for (auto v : adj[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++;
            APUtil(adj, v, visited, disc, low, time, u, isAP);
 
            // Check if the subtree rooted with v has
            // a connection to one of the ancestors of u
            low[u] = min(low[u], low[v]);
 
            // If u is not root and low value of one of
            // its child is more than discovery value of u.
            if (parent != -1 && low[v] >= disc[u])
                isAP[u] = true;
        }
 
        // Update low value of u for parent function calls.
        else if (v != parent)
            low[u] = min(low[u], disc[v]);
    }
 
    // If u is root of DFS tree and has two or more children.
    if (parent == -1 && children > 1)
        isAP[u] = true;
}
 
void AP(vector<int> adj[], int V)
{
    int disc[V] = { 0 };
    int low[V];
    bool visited[V] = { false };
    bool isAP[V] = { false };
    int time = 0, par = -1;
 
    // Adding this loop so that the
    // code works even if we are given
    // disconnected graph
    for (int u = 0; u < V; u++)
        if (!visited[u])
            APUtil(adj, u, visited, disc, low,
                   time, par, isAP);
 
    // Printing the APs
    for (int u = 0; u < V; u++)
        if (isAP[u] == true)
            cout << u << " ";
}
 
// Utility function to add an edge
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
 
int main()
{
    // Create graphs given in above diagrams
    cout << "Articulation points in first graph \n";
    int V = 5;
    vector<int> adj1[V];
    addEdge(adj1, 1, 0);
    addEdge(adj1, 0, 2);
    addEdge(adj1, 2, 1);
    addEdge(adj1, 0, 3);
    addEdge(adj1, 3, 4);
    AP(adj1, V);
 
    cout << "\nArticulation points in second graph \n";
    V = 4;
    vector<int> adj2[V];
    addEdge(adj2, 0, 1);
    addEdge(adj2, 1, 2);
    addEdge(adj2, 2, 3);
    AP(adj2, V);
 
    cout << "\nArticulation points in third graph \n";
    V = 7;
    vector<int> adj3[V];
    addEdge(adj3, 0, 1);
    addEdge(adj3, 1, 2);
    addEdge(adj3, 2, 0);
    addEdge(adj3, 1, 3);
    addEdge(adj3, 1, 4);
    addEdge(adj3, 1, 6);
    addEdge(adj3, 3, 5);
    addEdge(adj3, 4, 5);
    AP(adj3, V);
 
    return 0;
}


Java




// A Java program to find articulation
// points in an undirected graph
import java.util.*;
 
class Graph {
 
    static int time;
 
    static void addEdge(ArrayList<ArrayList<Integer> > adj, int u, int v)
    {
        adj.get(u).add(v);
        adj.get(v).add(u);
    }
 
    static void APUtil(ArrayList<ArrayList<Integer> > adj, int u,
                       boolean visited[], int disc[], int low[],
                       int parent, boolean isAP[])
    {
        // 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 adjacent to this
        for (Integer v : adj.get(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++;
                APUtil(adj, v, visited, disc, low, u, isAP);
 
                // Check if the subtree rooted with v has
                // a connection to one of the ancestors of u
                low[u] = Math.min(low[u], low[v]);
 
                // If u is not root and low value of one of
                // its child is more than discovery value of u.
                if (parent != -1 && low[v] >= disc[u])
                    isAP[u] = true;
            }
 
            // Update low value of u for parent function calls.
            else if (v != parent)
                low[u] = Math.min(low[u], disc[v]);
        }
 
        // If u is root of DFS tree and has two or more children.
        if (parent == -1 && children > 1)
            isAP[u] = true;
    }
 
    static void AP(ArrayList<ArrayList<Integer> > adj, int V)
    {
        boolean[] visited = new boolean[V];
        int[] disc = new int[V];
        int[] low = new int[V];
        boolean[] isAP = new boolean[V];
        int time = 0, par = -1;
 
        // Adding this loop so that the
        // code works even if we are given
        // disconnected graph
        for (int u = 0; u < V; u++)
            if (visited[u] == false)
                APUtil(adj, u, visited, disc, low, par, isAP);
 
        for (int u = 0; u < V; u++)
            if (isAP[u] == true)
                System.out.print(u + " ");
        System.out.println();
    }
 
    public static void main(String[] args)
    {
 
        // Creating first example graph
        int V = 5;
        ArrayList<ArrayList<Integer> > adj1 =
                         new ArrayList<ArrayList<Integer> >(V);
        for (int i = 0; i < V; i++)
            adj1.add(new ArrayList<Integer>());
        addEdge(adj1, 1, 0);
        addEdge(adj1, 0, 2);
        addEdge(adj1, 2, 1);
        addEdge(adj1, 0, 3);
        addEdge(adj1, 3, 4);
        System.out.println("Articulation points in first graph");
        AP(adj1, V);
 
        // Creating second example graph
        V = 4;
        ArrayList<ArrayList<Integer> > adj2 =
                         new ArrayList<ArrayList<Integer> >(V);
        for (int i = 0; i < V; i++)
            adj2.add(new ArrayList<Integer>());
 
        addEdge(adj2, 0, 1);
        addEdge(adj2, 1, 2);
        addEdge(adj2, 2, 3);
 
        System.out.println("Articulation points in second graph");
        AP(adj2, V);
 
        // Creating third example graph
        V = 7;
        ArrayList<ArrayList<Integer> > adj3 =
                            new ArrayList<ArrayList<Integer> >(V);
        for (int i = 0; i < V; i++)
            adj3.add(new ArrayList<Integer>());
 
        addEdge(adj3, 0, 1);
        addEdge(adj3, 1, 2);
        addEdge(adj3, 2, 0);
        addEdge(adj3, 1, 3);
        addEdge(adj3, 1, 4);
        addEdge(adj3, 1, 6);
        addEdge(adj3, 3, 5);
        addEdge(adj3, 4, 5);
 
        System.out.println("Articulation points in third graph");
 
        AP(adj3, V);
    }
}


Python3




# Python program to find articulation points in an undirected graph
  
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 find articulation points
    using DFS traversal
    u --> The vertex to be visited next
    visited[] --> keeps track of visited vertices
    disc[] --> Stores discovery times of visited vertices
    parent[] --> Stores parent vertices in DFS tree
    ap[] --> Store articulation points'''
    def APUtil(self, u, visited, ap, 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
                self.APUtil(v, visited, ap, parent, low, disc)
 
                # 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 children.
                if parent[u] == -1 and 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] != -1 and low[v] >= disc[u]:
                    ap[u] = True   
                     
                # Update low value of u for parent function calls   
            elif v != parent[u]:
                low[u] = min(low[u], disc[v])
 
 
    # The function to do DFS traversal. It uses recursive APUtil()
    def AP(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)
        ap = [False] * (self.V) # To store articulation points
 
        # Call the recursive helper function
        # to find articulation points
        # in DFS tree rooted with vertex 'i'
        for i in range(self.V):
            if visited[i] == False:
                self.APUtil(i, visited, ap, parent, low, disc)
 
        for index, value in enumerate (ap):
            if value == True: print (index,end=" ")
 
 # Create a graph given in the above diagram
g1 = Graph(5)
g1.addEdge(1, 0)
g1.addEdge(0, 2)
g1.addEdge(2, 1)
g1.addEdge(0, 3)
g1.addEdge(3, 4)
  
print ("\nArticulation points in first graph ")
g1.AP()
 
g2 = Graph(4)
g2.addEdge(0, 1)
g2.addEdge(1, 2)
g2.addEdge(2, 3)
print ("\nArticulation points in second graph ")
g2.AP()
 
  
g3 = Graph (7)
g3.addEdge(0, 1)
g3.addEdge(1, 2)
g3.addEdge(2, 0)
g3.addEdge(1, 3)
g3.addEdge(1, 4)
g3.addEdge(1, 6)
g3.addEdge(3, 5)
g3.addEdge(4, 5)
print ("\nArticulation points in third graph ")
g3.AP()
 
# This code is contributed by Neelam Yadav


C#




// A C# program to find articulation
// points in an undirected graph
using System;
using System.Collections.Generic;
 
// This class represents an undirected graph
// using adjacency list representation
public class Graph {
    private int V; // No. of vertices
 
    // Array of lists for Adjacency List Representation
    private List<int>[] adj;
    int time = 0;
    static readonly int NIL = -1;
 
    // Constructor
    Graph(int v)
    {
        V = v;
        adj = new List<int>[v];
        for (int i = 0; i < v; ++i)
            adj[i] = new List<int>();
    }
 
    // Function to add an edge into the graph
    void addEdge(int v, int w)
    {
        adj[v].Add(w); // Add w to v's list.
        adj[w].Add(v); // Add v to w's list
    }
 
    // A recursive function that find articulation points using DFS
    // u --> The vertex to be visited next
    // visited[] --> keeps track of visited vertices
    // disc[] --> Stores discovery times of visited vertices
    // parent[] --> Stores parent vertices in DFS tree
    // ap[] --> Store articulation points
    void APUtil(int u, bool[] visited, int[] disc,
                int[] low, int[] parent, bool[] ap)
    {
 
        // 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 adjacent to this
        foreach(int i in adj[u])
        {
            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] = Math.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 children.
                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] = Math.Min(low[u], disc[v]);
        }
    }
 
    // The function to do DFS traversal.
    // It uses recursive function APUtil()
    void 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)
                Console.Write(i + " ");
    }
 
    // Driver method
    public static void Main(String[] args)
    {
        // Create graphs given in above diagrams
        Console.WriteLine("Articulation points in first graph ");
        Graph g1 = new Graph(5);
        g1.addEdge(1, 0);
        g1.addEdge(0, 2);
        g1.addEdge(2, 1);
        g1.addEdge(0, 3);
        g1.addEdge(3, 4);
        g1.AP();
        Console.WriteLine();
 
        Console.WriteLine("Articulation points in Second graph");
        Graph g2 = new Graph(4);
        g2.addEdge(0, 1);
        g2.addEdge(1, 2);
        g2.addEdge(2, 3);
        g2.AP();
        Console.WriteLine();
 
        Console.WriteLine("Articulation points in Third graph ");
        Graph g3 = new Graph(7);
        g3.addEdge(0, 1);
        g3.addEdge(1, 2);
        g3.addEdge(2, 0);
        g3.addEdge(1, 3);
        g3.addEdge(1, 4);
        g3.addEdge(1, 6);
        g3.addEdge(3, 5);
        g3.addEdge(4, 5);
        g3.AP();
    }
}
 
// This code is contributed by PrinciRaj1992


Javascript




<script>
// A Javascript program to find articulation points in an undirected graph
 
// This class represents an undirected graph using adjacency list
// representation
class Graph
{
    // Constructor
    constructor(v)
    {
        this.V = v;
        this.adj = new Array(v);
        this.NIL = -1;
        this.time = 0;
        for (let i=0; i<v; ++i)
            this.adj[i] = [];
    }
     
    //Function to add an edge into the graph
    addEdge(v, w)
    {
        this.adj[v].push(w);  // Add w to v's list.
        this.adj[w].push(v);    //Add v to w's list
    }
     
    // A recursive function that find articulation points using DFS
    // u --> The vertex to be visited next
    // visited[] --> keeps track of visited vertices
    // disc[] --> Stores discovery times of visited vertices
    // parent[] --> Stores parent vertices in DFS tree
    // ap[] --> Store articulation points
    APUtil(u, visited, disc, low, parent, ap)
    {
        // Count of children in DFS Tree
        let children = 0;
  
        // Mark the current node as visited
        visited[u] = true;
  
        // Initialize discovery time and low value
        disc[u] = low[u] = ++this.time;
  
        // Go through all vertices adjacent to this
         
        for(let i of this.adj[u])
        {
            let 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;
                this.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]  = Math.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 children.
                if (parent[u] == this.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] != this.NIL && low[v] >= disc[u])
                    ap[u] = true;
            }
  
            // Update low value of u for parent function calls.
            else if (v != parent[u])
                low[u]  = Math.min(low[u], disc[v]);
        }
    }
     
    // The function to do DFS traversal. It uses recursive function APUtil()
    AP()
    {
        // Mark all the vertices as not visited
        let visited = new Array(this.V);
        let disc = new Array(this.V);
        let low = new Array(this.V);
        let parent = new Array(this.V);
        let ap = new Array(this.V); // To store articulation points
  
        // Initialize parent and visited, and ap(articulation point)
        // arrays
        for (let i = 0; i < this.V; i++)
        {
            parent[i] = this.NIL;
            visited[i] = false;
            ap[i] = false;
        }
  
        // Call the recursive helper function to find articulation
        // points in DFS tree rooted with vertex 'i'
        for (let i = 0; i < this.V; i++)
            if (visited[i] == false)
                this.APUtil(i, visited, disc, low, parent, ap);
  
        // Now ap[] contains articulation points, print them
        for (let i = 0; i < this.V; i++)
            if (ap[i] == true)
                document.write(i+" ");
    }
}
 
// Driver method
// Create graphs given in above diagrams
document.write("Articulation points in first graph  <br>");
let g1 = new Graph(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.AP();
document.write("<br>");
 
document.write("Articulation points in Second graph <br>");
let g2 = new Graph(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.AP();
document.write("<br>");
 
document.write("Articulation points in Third graph  <br>");
let g3 = new Graph(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.AP();
 
 
// This code is contributed by avanitrachhadiya2155
</script>


Output: 

Articulation points in first graph
0 3
Articulation points in second graph
1 2
Articulation points in third graph
1

Time Complexity: The above function is simple DFS with additional arrays. So time complexity is same as DFS which is O(V+E) for adjacency list representation of graph.

References: 
https://www.cs.washington.edu/education/courses/421/04su/slides/artic.pdf 
http://www.slideshare.net/TraianRebedea/algorithm-design-and-complexity-course-8 
http://faculty.simpson.edu/lydia.sinapova/www/cmsc250/LN250_Weiss/L25-Connectivity.htm
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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