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Biconnected Components
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Biconnected Components

Last Updated : 13 Feb, 2023
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A biconnected component is a maximal biconnected subgraph.

Biconnected Graph is already discussed here. In this article, we will see how to find biconnected component in a graph using algorithm by John Hopcroft and Robert Tarjan.

Biconnected Components

In above graph, following are the biconnected components: 

  • 4--2 3--4 3--1 2--3 1--2
  • 8--9
  • 8--5 7--8 5--7
  • 6--0 5--6 1--5 0--1
  • 10--11

Algorithm is based on Disc and Low Values discussed in Strongly Connected Components Article.

Idea is to store visited edges in a stack while DFS on a graph and keep looking for Articulation Points (highlighted in above figure). As soon as an Articulation Point u is found, all edges visited while DFS from node u onwards will form one biconnected component. When DFS completes for one connected component, all edges present in stack will form a biconnected component. 

If there is no Articulation Point in graph, then graph is biconnected and so there will be one biconnected component which is the graph itself.

C++
// A C++ program to find biconnected components in a given undirected graph #include <iostream> #include <list> #include <stack> #define NIL -1 using namespace std; int count = 0; class Edge { public:     int u;     int v;     Edge(int u, int v); }; Edge::Edge(int u, int v) {     this->u = u;     this->v = v; }  // A class that represents an directed graph class Graph {     int V; // No. of vertices     int E; // No. of edges     list<int>* adj; // A dynamic array of adjacency lists      // A Recursive DFS based function used by BCC()     void BCCUtil(int u, int disc[], int low[],                  list<Edge>* st, int parent[]);  public:     Graph(int V); // Constructor     void addEdge(int v, int w); // function to add an edge to graph     void BCC(); // prints strongly connected components };  Graph::Graph(int V) {     this->V = V;     this->E = 0;     adj = new list<int>[V]; }  void Graph::addEdge(int v, int w) {     adj[v].push_back(w);     E++; }  // A recursive function that finds and prints strongly connected // components using DFS traversal // u --> The vertex to be visited next // 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 // *st -- >> To store visited edges void Graph::BCCUtil(int u, int disc[], int low[], list<Edge>* st,                     int parent[]) {     // A static variable is used for simplicity, we can avoid use     // of static variable by passing a pointer.     static int time = 0;      // Initialize discovery time and low value     disc[u] = low[u] = ++time;     int children = 0;      // Go through all vertices adjacent to this     list<int>::iterator i;     for (i = adj[u].begin(); i != adj[u].end(); ++i) {         int v = *i; // v is current adjacent of 'u'          // If v is not visited yet, then recur for it         if (disc[v] == -1) {             children++;             parent[v] = u;             // store the edge in stack             st->push_back(Edge(u, v));             BCCUtil(v, disc, low, st, parent);              // Check if the subtree rooted with 'v' has a             // connection to one of the ancestors of 'u'             // Case 1 -- per Strongly Connected Components Article             low[u] = min(low[u], low[v]);              // If u is an articulation point,             // pop all edges from stack till u -- v             if ((disc[u] == 1 && children > 1) || (disc[u] > 1 && low[v] >= disc[u])) {                 while (st->back().u != u || st->back().v != v) {                     cout << st->back().u << "--" << st->back().v << " ";                     st->pop_back();                 }                 cout << st->back().u << "--" << st->back().v;                 st->pop_back();                 cout << endl;                 count++;             }         }          // Update low value of 'u' only of 'v' is still in stack         // (i.e. it's a back edge, not cross edge).         // Case 2 -- per Strongly Connected Components Article         else if (v != parent[u]) {             low[u] = min(low[u], disc[v]);             if (disc[v] < disc[u]) {                 st->push_back(Edge(u, v));             }         }     } }  // The function to do DFS traversal. It uses BCCUtil() void Graph::BCC() {     int* disc = new int[V];     int* low = new int[V];     int* parent = new int[V];     list<Edge>* st = new list<Edge>[E];      // Initialize disc and low, and parent arrays     for (int i = 0; i < V; i++) {         disc[i] = NIL;         low[i] = NIL;         parent[i] = NIL;     }      for (int i = 0; i < V; i++) {         if (disc[i] == NIL)             BCCUtil(i, disc, low, st, parent);          int j = 0;         // If stack is not empty, pop all edges from stack         while (st->size() > 0) {             j = 1;             cout << st->back().u << "--" << st->back().v << " ";             st->pop_back();         }         if (j == 1) {             cout << endl;             count++;         }     } }  // Driver program to test above function int main() {     Graph g(12);     g.addEdge(0, 1);     g.addEdge(1, 0);     g.addEdge(1, 2);     g.addEdge(2, 1);     g.addEdge(1, 3);     g.addEdge(3, 1);     g.addEdge(2, 3);     g.addEdge(3, 2);     g.addEdge(2, 4);     g.addEdge(4, 2);     g.addEdge(3, 4);     g.addEdge(4, 3);     g.addEdge(1, 5);     g.addEdge(5, 1);     g.addEdge(0, 6);     g.addEdge(6, 0);     g.addEdge(5, 6);     g.addEdge(6, 5);     g.addEdge(5, 7);     g.addEdge(7, 5);     g.addEdge(5, 8);     g.addEdge(8, 5);     g.addEdge(7, 8);     g.addEdge(8, 7);     g.addEdge(8, 9);     g.addEdge(9, 8);     g.addEdge(10, 11);     g.addEdge(11, 10);     g.BCC();     cout << "Above are " << count << " biconnected components in graph";     return 0; } 
Java
// A Java program to find biconnected components in a given // undirected graph import java.io.*; import java.util.*;  // This class represents a directed graph using adjacency // list representation class Graph {     private int V, E; // No. of vertices & Edges respectively     private LinkedList<Integer> adj[]; // Adjacency List      // Count is number of biconnected components. time is     // used to find discovery times     static int count = 0, time = 0;      class Edge {         int u;         int v;         Edge(int u, int v)         {             this.u = u;             this.v = v;         }     };      // Constructor     Graph(int v)     {         V = v;         E = 0;         adj = new LinkedList[v];         for (int i = 0; i < v; ++i)             adj[i] = new LinkedList();     }      // Function to add an edge into the graph     void addEdge(int v, int w)     {         adj[v].add(w);         E++;     }      // A recursive function that finds and prints strongly connected     // components using DFS traversal     // u --> The vertex to be visited next     // 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     // *st -- >> To store visited edges     void BCCUtil(int u, int disc[], int low[], LinkedList<Edge> st,                  int parent[])     {          // Initialize discovery time and low value         disc[u] = low[u] = ++time;         int children = 0;          // Go through all vertices adjacent to this         Iterator<Integer> it = adj[u].iterator();         while (it.hasNext()) {             int v = it.next(); // v is current adjacent of 'u'              // If v is not visited yet, then recur for it             if (disc[v] == -1) {                 children++;                 parent[v] = u;                  // store the edge in stack                 st.add(new Edge(u, v));                 BCCUtil(v, disc, low, st, parent);                  // Check if the subtree rooted with 'v' has a                 // connection to one of the ancestors of 'u'                 // Case 1 -- per Strongly Connected Components Article                 if (low[u] > low[v])                     low[u] = low[v];                  // If u is an articulation point,                 // pop all edges from stack till u -- v                 if ((disc[u] == 1 && children > 1) || (disc[u] > 1 && low[v] >= disc[u])) {                     while (st.getLast().u != u || st.getLast().v != v) {                         System.out.print(st.getLast().u + "--" + st.getLast().v + " ");                         st.removeLast();                     }                     System.out.println(st.getLast().u + "--" + st.getLast().v + " ");                     st.removeLast();                      count++;                 }             }              // Update low value of 'u' only if 'v' is still in stack             // (i.e. it's a back edge, not cross edge).             // Case 2 -- per Strongly Connected Components Article             else if (v != parent[u] && disc[v] < disc[u] ) {                 if (low[u] > disc[v])                     low[u] = disc[v];                  st.add(new Edge(u, v));             }         }     }      // The function to do DFS traversal. It uses BCCUtil()     void BCC()     {         int disc[] = new int[V];         int low[] = new int[V];         int parent[] = new int[V];         LinkedList<Edge> st = new LinkedList<Edge>();          // Initialize disc and low, and parent arrays         for (int i = 0; i < V; i++) {             disc[i] = -1;             low[i] = -1;             parent[i] = -1;         }          for (int i = 0; i < V; i++) {             if (disc[i] == -1)                 BCCUtil(i, disc, low, st, parent);              int j = 0;              // If stack is not empty, pop all edges from stack             while (st.size() > 0) {                 j = 1;                 System.out.print(st.getLast().u + "--" + st.getLast().v + " ");                 st.removeLast();             }             if (j == 1) {                 System.out.println();                 count++;             }         }     }      public static void main(String args[])     {         Graph g = new Graph(12);         g.addEdge(0, 1);         g.addEdge(1, 0);         g.addEdge(1, 2);         g.addEdge(2, 1);         g.addEdge(1, 3);         g.addEdge(3, 1);         g.addEdge(2, 3);         g.addEdge(3, 2);         g.addEdge(2, 4);         g.addEdge(4, 2);         g.addEdge(3, 4);         g.addEdge(4, 3);         g.addEdge(1, 5);         g.addEdge(5, 1);         g.addEdge(0, 6);         g.addEdge(6, 0);         g.addEdge(5, 6);         g.addEdge(6, 5);         g.addEdge(5, 7);         g.addEdge(7, 5);         g.addEdge(5, 8);         g.addEdge(8, 5);         g.addEdge(7, 8);         g.addEdge(8, 7);         g.addEdge(8, 9);         g.addEdge(9, 8);         g.addEdge(10, 11);         g.addEdge(11, 10);          g.BCC();          System.out.println("Above are " + g.count + " biconnected components in graph");     } } // This code is contributed by Aakash Hasija 
Python3
# Python program to find biconnected components in a given # undirected graph # Complexity : O(V + E)    from collections import defaultdict   # This class represents an directed graph  # using adjacency list representation class Graph:       def __init__(self, vertices):         # No. of vertices         self.V = vertices                   # default dictionary to store graph         self.graph = defaultdict(list)                  # time is used to find discovery times         self.Time = 0                   # Count is number of biconnected components         self.count = 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 finds and prints strongly connected     components using DFS traversal     u --> The vertex to be visited next     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     st -- >> To store visited edges'''     def BCCUtil(self, u, parent, low, disc, st):          # Count of children in current node          children = 0          # 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 disc[v] == -1 :                 parent[v] = u                 children += 1                 st.append((u, v)) # store the edge in stack                 self.BCCUtil(v, parent, low, disc, st)                  # Check if the subtree rooted with v has a connection to                 # one of the ancestors of u                 # Case 1 -- per Strongly Connected Components Article                 low[u] = min(low[u], low[v])                  # If u is an articulation point, pop                  # all edges from stack till (u, v)                 if parent[u] == -1 and children > 1 or parent[u] != -1 and low[v] >= disc[u]:                     self.count += 1 # increment count                     w = -1                     while w != (u, v):                         w = st.pop()                         print(w,end=" ")                     print()                          elif v != parent[u] and low[u] > disc[v]:                 '''Update low value of 'u' only of 'v' is still in stack                 (i.e. it's a back edge, not cross edge).                 Case 2                  -- per Strongly Connected Components Article'''                  low[u] = min(low [u], disc[v])                      st.append((u, v))       # The function to do DFS traversal.      # It uses recursive BCCUtil()     def BCC(self):                  # Initialize disc and low, and parent arrays         disc = [-1] * (self.V)         low = [-1] * (self.V)         parent = [-1] * (self.V)         st = []          # Call the recursive helper function to          # find articulation points         # in DFS tree rooted with vertex 'i'         for i in range(self.V):             if disc[i] == -1:                 self.BCCUtil(i, parent, low, disc, st)              # If stack is not empty, pop all edges from stack             if st:                 self.count = self.count + 1                  while st:                     w = st.pop()                     print(w,end=" ")                 print ()  # Create a graph given in the above diagram  g = Graph(12) g.addEdge(0, 1) g.addEdge(1, 2) g.addEdge(1, 3) g.addEdge(2, 3) g.addEdge(2, 4) g.addEdge(3, 4) g.addEdge(1, 5) g.addEdge(0, 6) g.addEdge(5, 6) g.addEdge(5, 7) g.addEdge(5, 8) g.addEdge(7, 8) g.addEdge(8, 9) g.addEdge(10, 11)  g.BCC(); print ("Above are % d biconnected components in graph" %(g.count));  # This code is contributed by Neelam Yadav 
C#
// A C# program to find biconnected components in a given  // undirected graph  using System; using System.Collections.Generic;  // This class represents a directed graph using adjacency  // list representation  public class Graph  {      private int V, E; // No. of vertices & Edges respectively      private List<int> []adj; // Adjacency List       // Count is number of biconnected components. time is      // used to find discovery times      int count = 0, time = 0;       class Edge      {          public int u;          public int v;          public Edge(int u, int v)          {              this.u = u;              this.v = v;          }      };       // Constructor      public Graph(int v)      {          V = v;          E = 0;          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);          E++;      }       // A recursive function that finds and prints strongly connected      // components using DFS traversal      // u --> The vertex to be visited next      // 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      // *st -- >> To store visited edges      void BCCUtil(int u, int []disc, int []low, List<Edge> st,                  int []parent)      {           // Initialize discovery time and low value          disc[u] = low[u] = ++time;          int children = 0;           // Go through all vertices adjacent to this          foreach(int it in adj[u])         {              int v = it; // v is current adjacent of 'u'               // If v is not visited yet, then recur for it              if (disc[v] == -1)              {                  children++;                  parent[v] = u;                   // store the edge in stack                  st.Add(new Edge(u, v));                  BCCUtil(v, disc, low, st, parent);                   // Check if the subtree rooted with 'v' has a                  // connection to one of the ancestors of 'u'                  // Case 1 -- per Strongly Connected Components Article                  if (low[u] > low[v])                      low[u] = low[v];                   // If u is an articulation point,                  // pop all edges from stack till u -- v                  if ((disc[u] == 1 && children > 1) ||                      (disc[u] > 1 && low[v] >= disc[u]))                  {                      while (st[st.Count-1].u != u ||                             st[st.Count-1].v != v)                      {                          Console.Write(st[st.Count - 1].u + "--" +                                        st[st.Count - 1].v + " ");                          st.RemoveAt(st.Count - 1);                      }                      Console.WriteLine(st[st.Count - 1].u + "--" +                                        st[st.Count - 1].v + " ");                      st.RemoveAt(st.Count - 1);                       count++;                  }              }               // Update low value of 'u' only if 'v' is still in stack              // (i.e. it's a back edge, not cross edge).              // Case 2 -- per Strongly Connected Components Article              else if (v != parent[u] && disc[v] < disc[u] )              {                  if (low[u] > disc[v])                      low[u] = disc[v];                   st.Add(new Edge(u, v));              }          }      }       // The function to do DFS traversal. It uses BCCUtil()      void BCC()      {          int []disc = new int[V];          int []low = new int[V];          int []parent = new int[V];          List<Edge> st = new List<Edge>();           // Initialize disc and low, and parent arrays          for (int i = 0; i < V; i++)          {              disc[i] = -1;              low[i] = -1;              parent[i] = -1;          }           for (int i = 0; i < V; i++)         {              if (disc[i] == -1)                  BCCUtil(i, disc, low, st, parent);               int j = 0;               // If stack is not empty, pop all edges from stack              while (st.Count > 0)              {                  j = 1;                  Console.Write(st[st.Count - 1].u + "--" +                              st[st.Count - 1].v + " ");                  st.RemoveAt(st.Count - 1);              }              if (j == 1)              {                  Console.WriteLine();                  count++;              }          }      }       // Driver code     public static void Main(String []args)      {          Graph g = new Graph(12);          g.addEdge(0, 1);          g.addEdge(1, 0);          g.addEdge(1, 2);          g.addEdge(2, 1);          g.addEdge(1, 3);          g.addEdge(3, 1);          g.addEdge(2, 3);          g.addEdge(3, 2);          g.addEdge(2, 4);          g.addEdge(4, 2);          g.addEdge(3, 4);          g.addEdge(4, 3);          g.addEdge(1, 5);          g.addEdge(5, 1);          g.addEdge(0, 6);          g.addEdge(6, 0);          g.addEdge(5, 6);          g.addEdge(6, 5);          g.addEdge(5, 7);          g.addEdge(7, 5);          g.addEdge(5, 8);          g.addEdge(8, 5);          g.addEdge(7, 8);          g.addEdge(8, 7);          g.addEdge(8, 9);          g.addEdge(9, 8);          g.addEdge(10, 11);          g.addEdge(11, 10);           g.BCC();           Console.WriteLine("Above are " + g.count +                         " biconnected components in graph");      }  }  // This code is contributed by PrinciRaj1992 
JavaScript
<script> // A Javascript program to find biconnected components in a given // undirected graph  class Edge {     constructor(u,v)     {         this.u = u;             this.v = v;     } }  // This class represents a directed graph using adjacency // list representation class Graph {     // Constructor     constructor(v)     {         this.count=0;         this.time = 0;         this.V = v;         this.E = 0;         this.adj = new Array(v);         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);         this.E++;     }          // A recursive function that finds and prints strongly connected     // components using DFS traversal     // u --> The vertex to be visited next     // 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     // *st -- >> To store visited edges     BCCUtil(u,disc,low,st,parent)     {         // Initialize discovery time and low value         disc[u] = low[u] = ++this.time;         let children = 0;            // Go through all vertices adjacent to this                  for(let it of this.adj[u].values()) {             let v = it; // v is current adjacent of 'u'                // If v is not visited yet, then recur for it             if (disc[v] == -1) {                 children++;                 parent[v] = u;                    // store the edge in stack                 st.push(new Edge(u, v));                 this.BCCUtil(v, disc, low, st, parent);                    // Check if the subtree rooted with 'v' has a                 // connection to one of the ancestors of 'u'                 // Case 1 -- per Strongly Connected Components Article                 if (low[u] > low[v])                     low[u] = low[v];                    // If u is an articulation point,                 // pop all edges from stack till u -- v                 if ((disc[u] == 1 && children > 1) || (disc[u] > 1 && low[v] >= disc[u])) {                     while (st[st.length-1].u != u || st[st.length-1].v != v) {                         document.write(st[st.length-1].u + "--" + st[st.length-1].v + " ");                         st.pop();                     }                     document.write(st[st.length-1].u + "--" + st[st.length-1].v + " <br>");                     st.pop();                        this.count++;                 }             }                // Update low value of 'u' only if 'v' is still in stack             // (i.e. it's a back edge, not cross edge).             // Case 2 -- per Strongly Connected Components Article             else if (v != parent[u] && disc[v] < disc[u] ) {                 if (low[u] > disc[v])                     low[u] = disc[v];                    st.push(new Edge(u, v));             }         }     }          // The function to do DFS traversal. It uses BCCUtil()     BCC()     {         let disc = new Array(this.V);         let low = new Array(this.V);         let parent = new Array(this.V);         let st = [];            // Initialize disc and low, and parent arrays         for (let i = 0; i < this.V; i++) {             disc[i] = -1;             low[i] = -1;             parent[i] = -1;         }            for (let i = 0; i < this.V; i++) {             if (disc[i] == -1)                 this.BCCUtil(i, disc, low, st, parent);                let j = 0;                // If stack is not empty, pop all edges from stack             while (st.length > 0) {                 j = 1;                 document.write(st[st.length-1].u + "--" + st[st.length-1].v + " ");                 st.pop();             }             if (j == 1) {                 document.write("<br>");                 this.count++;             }     } } }  let g = new Graph(12); g.addEdge(0, 1); g.addEdge(1, 0); g.addEdge(1, 2); g.addEdge(2, 1); g.addEdge(1, 3); g.addEdge(3, 1); g.addEdge(2, 3); g.addEdge(3, 2); g.addEdge(2, 4); g.addEdge(4, 2); g.addEdge(3, 4); g.addEdge(4, 3); g.addEdge(1, 5); g.addEdge(5, 1); g.addEdge(0, 6); g.addEdge(6, 0); g.addEdge(5, 6); g.addEdge(6, 5); g.addEdge(5, 7); g.addEdge(7, 5); g.addEdge(5, 8); g.addEdge(8, 5); g.addEdge(7, 8); g.addEdge(8, 7); g.addEdge(8, 9); g.addEdge(9, 8); g.addEdge(10, 11); g.addEdge(11, 10);  g.BCC();  document.write("Above are " + g.count + " biconnected components in graph");  // This code is contributed by avanitrachhadiya2155 </script> 

Output
4--2 3--4 3--1 2--3 1--2 8--9 8--5 7--8 5--7 6--0 5--6 1--5 0--1  10--11 Above are 5 biconnected components in graph

Time Complexity : O(N + E), where N is the number of nodes and E is the number of edges, as the time complexity of Depth first search.
Space Complexity :O(N), here N is the number of nodes in the graph, we need a O(N) recursion stack space to make DFS calls. 


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Biconnected Components

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Article Tags :
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  • DSA
  • graph-connectivity
Practice Tags :
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    Graph is a non-linear data structure that contains nodes (vertices) and edges. A graph is a collection of set of vertices and edges (formed by connecting two vertices). A graph is defined as G = {V, E} where V is the set of vertices and E is the set of edges. Graphs can be used to model a wide varie
    7 min read
    Transpose graph
    Transpose of a directed graph G is another directed graph on the same set of vertices with all of the edges reversed compared to the orientation of the corresponding edges in G. That is, if G contains an edge (u, v) then the converse/transpose/reverse of G contains an edge (v, u) and vice versa. Giv
    9 min read
    Difference Between Graph and Tree
    Graphs and trees are two fundamental data structures used in computer science to represent relationships between objects. While they share some similarities, they also have distinct differences that make them suitable for different applications. Difference Between Graph and Tree What is Graph?A grap
    2 min read

    BFS and DFS on Graph

    Breadth First Search or BFS for a Graph
    Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
    15+ min read
    Depth First Search or DFS for a Graph
    In Depth First Search (or DFS) for a graph, we traverse all adjacent vertices one by one. When we traverse an adjacent vertex, we completely finish the traversal of all vertices reachable through that adjacent vertex. This is similar to a tree, where we first completely traverse the left subtree and
    13 min read
    Applications, Advantages and Disadvantages of Depth First Search (DFS)
    Depth First Search is a widely used algorithm for traversing a graph. Here we have discussed some applications, advantages, and disadvantages of the algorithm. Applications of Depth First Search:1. Detecting cycle in a graph: A graph has a cycle if and only if we see a back edge during DFS. So we ca
    4 min read
    Applications, Advantages and Disadvantages of Breadth First Search (BFS)
    We have earlier discussed Breadth First Traversal Algorithm for Graphs. Here in this article, we will see the applications, advantages, and disadvantages of the Breadth First Search. Applications of Breadth First Search: 1. Shortest Path and Minimum Spanning Tree for unweighted graph: In an unweight
    4 min read
    Iterative Depth First Traversal of Graph
    Given a directed Graph, the task is to perform Depth First Search of the given graph.Note: Start DFS from node 0, and traverse the nodes in the same order as adjacency list.Note : There can be multiple DFS traversals of a graph according to the order in which we pick adjacent vertices. Here we pick
    10 min read
    BFS for Disconnected Graph
    In the previous post, BFS only with a particular vertex is performed i.e. it is assumed that all vertices are reachable from the starting vertex. But in the case of a disconnected graph or any vertex that is unreachable from all vertex, the previous implementation will not give the desired output, s
    14 min read
    Transitive Closure of a Graph using DFS
    Given a directed graph, find out if a vertex v is reachable from another vertex u for all vertex pairs (u, v) in the given graph. Here reachable means that there is a path from vertex u to v. The reach-ability matrix is called transitive closure of a graph. For example, consider below graph: GraphTr
    8 min read
    Difference between BFS and DFS
    Breadth-First Search (BFS) and Depth-First Search (DFS) are two fundamental algorithms used for traversing or searching graphs and trees. This article covers the basic difference between Breadth-First Search and Depth-First Search.Difference between BFS and DFSParametersBFSDFSStands forBFS stands fo
    2 min read

    Cycle in a Graph

    Detect Cycle in a Directed Graph
    Given the number of vertices V and a list of directed edges, determine whether the graph contains a cycle or not.Examples: Input: V = 4, edges[][] = [[0, 1], [0, 2], [1, 2], [2, 0], [2, 3]]Cycle: 0 → 2 → 0 Output: trueExplanation: The diagram clearly shows a cycle 0 → 2 → 0 Input: V = 4, edges[][] =
    15+ min read
    Detect cycle in an undirected graph
    Given an undirected graph, the task is to check if there is a cycle in the given graph.Examples:Input: V = 4, edges[][]= [[0, 1], [0, 2], [1, 2], [2, 3]]Undirected Graph with 4 vertices and 4 edgesOutput: trueExplanation: The diagram clearly shows a cycle 0 → 2 → 1 → 0Input: V = 4, edges[][] = [[0,
    8 min read
    Detect Cycle in a directed graph using colors
    Given a directed graph represented by the number of vertices V and a list of directed edges, determine whether the graph contains a cycle.Your task is to implement a function that accepts V (number of vertices) and edges (an array of directed edges where each edge is a pair [u, v]), and returns true
    9 min read
    Detect a negative cycle in a Graph | (Bellman Ford)
    Given a directed weighted graph, your task is to find whether the given graph contains any negative cycles that are reachable from the source vertex (e.g., node 0).Note: A negative-weight cycle is a cycle in a graph whose edges sum to a negative value.Example:Input: V = 4, edges[][] = [[0, 3, 6], [1
    15+ min read
    Cycles of length n in an undirected and connected graph
    Given an undirected and connected graph and a number n, count the total number of simple cycles of length n in the graph. A simple cycle of length n is defined as a cycle that contains exactly n vertices and n edges. Note that for an undirected graph, each cycle should only be counted once, regardle
    10 min read
    Detecting negative cycle using Floyd Warshall
    We are given a directed graph. We need compute whether the graph has negative cycle or not. A negative cycle is one in which the overall sum of the cycle comes negative. Negative weights are found in various applications of graphs. For example, instead of paying cost for a path, we may get some adva
    12 min read
    Clone a Directed Acyclic Graph
    A directed acyclic graph (DAG) is a graph which doesn't contain a cycle and has directed edges. We are given a DAG, we need to clone it, i.e., create another graph that has copy of its vertices and edges connecting them. Examples: Input : 0 - - - > 1 - - - -> 4 | / \ ^ | / \ | | / \ | | / \ |
    12 min read
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