Minimum number of edges required to be removed from an Undirected Graph to make it acyclic
Given an undirected graph consisting of N nodes containing values from the range [1, N] and M edges in a matrix Edges[][], the task is to determine the minimum number of edges required to be removed such that the resulting graph does not contain any cycle.
Examples:
Input: N = 3, M = 3, edges[][] = [[1, 2], [2, 3], [3, 1]]
Output: 1
Explanation:
Removing any one of the edges will make the graph acyclic. Therefore, at least one edge needs to be removed.Input: N = 3, M = 2, edges[][] = [[1, 2], [2, 3]]
Output: 0
Explanation: Graph is already acyclic. Therefore, no edge removal is required.
Naive Approach: The simplest approach is to try deleting all possible combination of sequence of edges from the given graph one by one and for each combination, count the number of removals required to make the graph acyclic. Finally, among these combinations, choose the one which deletes the minimum number of edges to obtain an acyclic graph.
Time Complexity: O(M!)
Auxiliary Space: O(N + M)
Efficient Approach: The above approach can be optimized based on the following observations:
- A graph is acyclic when it is a Tree or a forest of trees(disconnected groups of trees).
- A tree with C nodes will have (C – 1) edges.
- If there are K connected components from C1 to CK, then minimum number of edges to be removed is equal to:
M – (C1 – 1) – (C2 – 1) … (Ck -1 )
=> M – (C1 + … + CK) + K
=> M – N + K
Follow the steps below to solve the problem:
- Find the number of connected components from the given graph using DFS.
- Considering the count of connected components to be K, then print M – N + K as the required minimum number of edges to be removed to make the resulting graph acyclic.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Stores the adjacency list vector< int > vec[100001]; // Stores if a vertex is // visited or not bool vis[100001]; int cc = 1; // Function to perform DFS Traversal // to count the number and size of // all connected components void dfs( int node) { // Mark the current node as visited vis[node] = true ; // Traverse the adjacency list // of the current node for ( auto x : vec[node]) { // For every unvisited node if (!vis[x]) { cc++; // Recursive DFS Call dfs(x); } } } // Function to add undirected // edge in the graph void addEdge( int u, int v) { vec[u].push_back(v); vec[v].push_back(u); } // Function to calculate minimum // number of edges to be removed void minEdgeRemoved( int N, int M, int Edges[][2]) { // Create Adjacency list for ( int i = 0; i < M; i++) { addEdge(Edges[i][0], Edges[i][1]); } memset (vis, false , sizeof (vis)); int k = 0; // Iterate over all the nodes for ( int i = 1; i <= N; i++) { if (!vis[i]) { cc = 1; dfs(i); k++; } } // Print the final count cout << M - N + k << endl; } // Driver Code int main() { int N = 3, M = 2; int Edges[][2] = { { 1, 2 }, { 2, 3 } }; minEdgeRemoved(N, M, Edges); } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Stores the adjacency list @SuppressWarnings ( "unchecked" ) static Vector<Integer> []vec = new Vector[ 100001 ]; // Stores if a vertex is // visited or not static boolean []vis = new boolean [ 100001 ]; static int cc = 1 ; // Function to perform DFS Traversal // to count the number and size of // all connected components static void dfs( int node) { // Mark the current node as visited vis[node] = true ; // Traverse the adjacency list // of the current node for ( int x : vec[node]) { // For every unvisited node if (!vis[x]) { cc++; // Recursive DFS call dfs(x); } } } // Function to add undirected // edge in the graph static void addEdge( int u, int v) { vec[u].add(v); vec[v].add(u); } // Function to calculate minimum // number of edges to be removed static void minEdgeRemoved( int N, int M, int Edges[][]) { // Create Adjacency list for ( int i = 0 ; i < M; i++) { addEdge(Edges[i][ 0 ], Edges[i][ 1 ]); } int k = 0 ; // Iterate over all the nodes for ( int i = 1 ; i <= N; i++) { if (!vis[i]) { cc = 1 ; dfs(i); k++; } } // Print the final count System.out.print(M - N + k + "\n" ); } // Driver Code public static void main(String[] args) { int N = 3 , M = 2 ; int Edges[][] = { { 1 , 2 }, { 2 , 3 } }; for ( int i = 0 ; i < vec.length; i++) vec[i] = new Vector<Integer>(); minEdgeRemoved(N, M, Edges); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement # the above approach # Stores the adjacency list vec = [[] for i in range ( 100001 )] # Stores if a vertex is # visited or not vis = [ False ] * 100001 cc = 1 # Function to perform DFS Traversal # to count the number and size of # all connected components def dfs(node): global cc # Mark the current node as visited vis[node] = True # Traverse the adjacency list # of the current node for x in vec[node]: # For every unvisited node if (vis[x] = = 0 ): cc + = 1 # Recursive DFS Call dfs(x) # Function to add undirected # edge in the graph def addEdge(u, v): vec[u].append(v) vec[v].append(u) # Function to calculate minimum # number of edges to be removed def minEdgeRemoved(N, M, Edges): global cc # Create Adjacency list for i in range (M): addEdge(Edges[i][ 0 ], Edges[i][ 1 ]) # memset(vis, false, sizeof(vis)) k = 0 # Iterate over all the nodes for i in range ( 1 , N + 1 ): if ( not vis[i]): cc = 1 dfs(i) k + = 1 # Print the final count print (M - N + k) # Driver Code if __name__ = = '__main__' : N = 3 M = 2 Edges = [ [ 1 , 2 ], [ 2 , 3 ] ] minEdgeRemoved(N, M, Edges) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Stores the adjacency list static List< int > []vec = new List< int >[100001]; // Stores if a vertex is // visited or not static bool []vis = new bool [100001]; static int cc = 1; // Function to perform DFS Traversal // to count the number and size of // all connected components static void dfs( int node) { // Mark the current node as visited vis[node] = true ; // Traverse the adjacency list // of the current node foreach ( int x in vec[node]) { // For every unvisited node if (!vis[x]) { cc++; // Recursive DFS call dfs(x); } } } // Function to add undirected // edge in the graph static void addEdge( int u, int v) { vec[u].Add(v); vec[v].Add(u); } // Function to calculate minimum // number of edges to be removed static void minEdgeRemoved( int N, int M, int [,]Edges) { // Create Adjacency list for ( int i = 0; i < M; i++) { addEdge(Edges[i, 0], Edges[i, 1]); } int k = 0; // Iterate over all the nodes for ( int i = 1; i <= N; i++) { if (!vis[i]) { cc = 1; dfs(i); k++; } } // Print the readonly count Console.Write(M - N + k + "\n" ); } // Driver Code public static void Main(String[] args) { int N = 3, M = 2; int [,]Edges = { { 1, 2 }, { 2, 3 } }; for ( int i = 0; i < vec.Length; i++) vec[i] = new List< int >(); minEdgeRemoved(N, M, Edges); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of the above approach // Stores the adjacency list let vec = new Array(100001); // Stores if a vertex is // visited or not let vis = new Array(100001); vis.fill( false ); let cc = 1; // Function to perform DFS Traversal // to count the number and size of // all connected components function dfs(node) { // Mark the current node as visited vis[node] = true ; // Traverse the adjacency list // of the current node for (let x = 0; x < vec[node].length; x++) { // For every unvisited node if (!vis[vec[node][x]]) { cc++; // Recursive DFS call dfs(vec[node][x]); } } } // Function to add undirected // edge in the graph function addEdge(u, v) { vec[u].push(v); vec[v].push(u); } // Function to calculate minimum // number of edges to be removed function minEdgeRemoved(N, M, Edges) { // Create Adjacency list for (let i = 0; i < M; i++) { addEdge(Edges[i][0], Edges[i][1]); } let k = 0; // Iterate over all the nodes for (let i = 1; i <= N; i++) { if (!vis[i]) { cc = 1; dfs(i); k++; } } // Print the readonly count document.write((M - N + k) + "</br>" ); } let N = 3, M = 2; let Edges = [ [ 1, 2 ], [ 2, 3 ] ]; for (let i = 0; i < vec.length; i++) vec[i] = []; minEdgeRemoved(N, M, Edges); </script> |
0
Time Complexity: O(N + M)
Auxiliary Space: O(N + M)