Vertex Cover Problem (Dynamic Programming Solution for Tree)
A vertex cover of an undirected graph is a subset of its vertices such that for every edge (u, v) of the graph, either ‘u’ or ‘v’ is in vertex cover. Although the name is Vertex Cover, the set covers all edges of the given graph.
The problem to find minimum size vertex cover of a graph is NP complete. But it can be solved in polynomial time for trees. In this post a solution for Binary Tree is discussed. The same solution can be extended for n-ary trees.
For example, consider the following binary tree. The smallest vertex cover is {20, 50, 30} and size of the vertex cover is 3.
The idea is to consider following two possibilities for root and recursively for all nodes down the root.
1) Root is part of vertex cover: In this case root covers all children edges. We recursively calculate size of vertex covers for left and right subtrees and add 1 to the result (for root).
2) Root is not part of vertex cover: In this case, both children of root must be included in vertex cover to cover all root to children edges. We recursively calculate size of vertex covers of all grandchildren and number of children to the result (for two children of root).
Below are implementation of above idea.
C
// A naive recursive C implementation for vertex cover problem for a tree#include <stdio.h>#include <stdlib.h>// A utility function to find min of two integersint min(int x, int y) { return (x < y)? x: y; }/* A binary tree node has data, pointer to left child and a pointer to right child */struct node{ int data; struct node *left, *right;};// The function returns size of the minimum vertex coverint vCover(struct node *root){ // The size of minimum vertex cover is zero if tree is empty or there // is only one node if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return 0; // Calculate size of vertex cover when root is part of it int size_incl = 1 + vCover(root->left) + vCover(root->right); // Calculate size of vertex cover when root is not part of it int size_excl = 0; if (root->left) size_excl += 1 + vCover(root->left->left) + vCover(root->left->right); if (root->right) size_excl += 1 + vCover(root->right->left) + vCover(root->right->right); // Return the minimum of two sizes return min(size_incl, size_excl);}// A utility function to create a nodestruct node* newNode( int data ){ struct node* temp = (struct node *) malloc( sizeof(struct node) ); temp->data = data; temp->left = temp->right = NULL; return temp;}// Driver program to test above functionsint main(){ // Let us construct the tree given in the above diagram struct node *root = newNode(20); root->left = newNode(8); root->left->left = newNode(4); root->left->right = newNode(12); root->left->right->left = newNode(10); root->left->right->right = newNode(14); root->right = newNode(22); root->right->right = newNode(25); printf ("Size of the smallest vertex cover is %d ", vCover(root)); return 0;} |
C++
#include <iostream>#include <cstdlib>// A utility function to find min of two integersint min(int x, int y) { return (x < y)? x: y; }/* A binary tree node has data, pointer to left child and a pointer to right child */struct node{ int data; node *left, *right;};// The function returns size of the minimum vertex coverint vCover(node *root){ // The size of minimum vertex cover is zero if tree is empty or there // is only one node if (root == nullptr) return 0; if (root->left == nullptr && root->right == nullptr) return 0; // Calculate size of vertex cover when root is part of it int size_incl = 1 + vCover(root->left) + vCover(root->right); // Calculate size of vertex cover when root is not part of it int size_excl = 0; if (root->left) size_excl += 1 + vCover(root->left->left) + vCover(root->left->right); if (root->right) size_excl += 1 + vCover(root->right->left) + vCover(root->right->right); // Return the minimum of two sizes return min(size_incl, size_excl);}// A utility function to create a nodenode* newNode( int data ){ node* temp = new node; temp->data = data; temp->left = temp->right = nullptr; return temp;}// Driver program to test above functionsint main(){ // Let us construct the tree given in the above diagram node *root = newNode(20); root->left = newNode(8); root->left->left = newNode(4); root->left->right = newNode(12); root->left->right->left = newNode(10); root->left->right->right = newNode(14); root->right = newNode(22); root->right->right = newNode(25); std::cout << "Size of the smallest vertex cover is " << vCover(root) << std::endl; return 0;} |
Java
// A naive recursive Java implementation // for vertex cover problem for a treeclass GFG{ // A utility function to find min of two integers static int min(int x, int y) { return (x < y) ? x : y; } /* * A binary tree node has data, pointer to left child and a pointer to right * child */ static class node { int data; node left, right; }; // The function returns size // of the minimum vertex cover static int vCover(node root) { // The size of minimum vertex cover // is zero if tree is empty or there // is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // Calculate size of vertex cover // when root is part of it int size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it int size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Return the minimum of two sizes return Math.min(size_incl, size_excl); } // A utility function to create a node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = temp.right = null; return temp; } // Driver code public static void main(String[] args) { // Let us construct tree given in the above diagram node root = newNode(20); root.left = newNode(8); root.left.left = newNode(4); root.left.right = newNode(12); root.left.right.left = newNode(10); root.left.right.right = newNode(14); root.right = newNode(22); root.right.right = newNode(25); System.out.printf("Size of the smallest vertex" + "cover is %d ", vCover(root)); }}// This code is contributed by 29AjayKumar |
Python3
# A naive recursive Python3 implementation # for vertex cover problem for a tree# A utility function to find min of two integers# A binary tree node has data, pointer to# left child and a pointer to right child class Node: def __init__(self, x): self.data = x self.left = None self.right = None# The function returns size of # the minimum vertex coverdef vCover(root): # The size of minimum vertex cover # is zero if tree is empty or there # is only one node if (root == None): return 0 if (root.left == None and root.right == None): return 0 # Calculate size of vertex cover when # root is part of it size_incl = (1 + vCover(root.left) + vCover(root.right)) # Calculate size of vertex cover # when root is not part of it size_excl = 0 if (root.left): size_excl += (1 + vCover(root.left.left) + vCover(root.left.right)) if (root.right): size_excl += (1 + vCover(root.right.left) + vCover(root.right.right)) # Return the minimum of two sizes return min(size_incl, size_excl)# Driver Codeif __name__ == '__main__': # Let us construct the tree # given in the above diagram root = Node(20) root.left = Node(8) root.left.left = Node(4) root.left.right = Node(12) root.left.right.left = Node(10) root.left.right.right = Node(14) root.right = Node(22) root.right.right = Node(25) print("Size of the smallest vertex cover is", vCover(root))# This code is contributed by mohit kumar 29 |
C#
// A naive recursive C# implementation // for vertex cover problem for a treeusing System;class GFG{ // A utility function to find // min of two integers static int min(int x, int y) { return (x < y) ? x : y; } /* * A binary tree node has data, pointer to left child and a pointer to right * child */ public class node { public int data; public node left, right; }; // The function returns size // of the minimum vertex cover static int vCover(node root) { // The size of minimum vertex cover // is zero if tree is empty or there // is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // Calculate size of vertex cover // when root is part of it int size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it int size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Return the minimum of two sizes return Math.Min(size_incl, size_excl); } // A utility function to create a node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = temp.right = null; return temp; } // Driver code public static void Main(String[] args) { // Let us construct tree given // in the above diagram node root = newNode(20); root.left = newNode(8); root.left.left = newNode(4); root.left.right = newNode(12); root.left.right.left = newNode(10); root.left.right.right = newNode(14); root.right = newNode(22); root.right.right = newNode(25); Console.Write("Size of the smallest vertex" + "cover is {0} ", vCover(root)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// A naive recursive Javascript implementation// for vertex cover problem for a tree // A utility function to find min of two integers function min(x,y) { return (x < y) ? x : y; } /* * A binary tree node has data, pointer to left child and a pointer to right * child */ class Node { constructor(d) { this.data=d; this.left=null; this.right=null; } } // The function returns size // of the minimum vertex cover function vCover(root) { // The size of minimum vertex cover // is zero if tree is empty or there // is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // Calculate size of vertex cover // when root is part of it let size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it let size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Return the minimum of two sizes return Math.min(size_incl, size_excl); } // Driver code // Let us construct tree given in the above diagram root = new Node(20); root.left = new Node(8); root.left.left = new Node(4); root.left.right = new Node(12); root.left.right.left = new Node(10); root.left.right.right = new Node(14); root.right = new Node(22); root.right.right = new Node(25); document.write("Size of the smallest vertex" + "cover is ", vCover(root)); // This code is contributed by unknown2108</script> |
Output
Size of the smallest vertex cover is 3
Time complexity of the above naive recursive approach is exponential. It should be noted that the above function computes the same subproblems again and again. For example, vCover of node with value 50 is evaluated twice as 50 is grandchild of 10 and child of 20.
Since same subproblems are called again, this problem has Overlapping Subproblems property. So Vertex Cover problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computations of same subproblems can be avoided by storing the solutions to subproblems and solving problems in bottom up manner.
Following is the implementation of Dynamic Programming based solution. In the following solution, an additional field ‘vc’ is added to tree nodes. The initial value of ‘vc’ is set as 0 for all nodes. The recursive function vCover() calculates ‘vc’ for a node only if it is not already set.
C
/* Dynamic programming based program for Vertex Cover problem for a Binary Tree */#include <stdio.h>#include <stdlib.h>// A utility function to find min of two integersint min(int x, int y) { return (x < y)? x: y; }/* A binary tree node has data, pointer to left child and a pointer to right child */struct node{ int data; int vc; struct node *left, *right;};// A memoization based function that returns size of the minimum vertex cover.int vCover(struct node *root){ // The size of minimum vertex cover is zero if tree is empty or there // is only one node if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return 0; // If vertex cover for this node is already evaluated, then return it // to save recomputation of same subproblem again. if (root->vc != 0) return root->vc; // Calculate size of vertex cover when root is part of it int size_incl = 1 + vCover(root->left) + vCover(root->right); // Calculate size of vertex cover when root is not part of it int size_excl = 0; if (root->left) size_excl += 1 + vCover(root->left->left) + vCover(root->left->right); if (root->right) size_excl += 1 + vCover(root->right->left) + vCover(root->right->right); // Minimum of two values is vertex cover, store it before returning root->vc = min(size_incl, size_excl); return root->vc;}// A utility function to create a nodestruct node* newNode( int data ){ struct node* temp = (struct node *) malloc( sizeof(struct node) ); temp->data = data; temp->left = temp->right = NULL; temp->vc = 0; // Set the vertex cover as 0 return temp;}// Driver program to test above functionsint main(){ // Let us construct the tree given in the above diagram struct node *root = newNode(20); root->left = newNode(8); root->left->left = newNode(4); root->left->right = newNode(12); root->left->right->left = newNode(10); root->left->right->right = newNode(14); root->right = newNode(22); root->right->right = newNode(25); printf ("Size of the smallest vertex cover is %d ", vCover(root)); return 0;} |
C++
#include <iostream>#include <stdlib.h>// A utility function to find min of two integersint min(int x, int y) { return (x < y)? x: y; }/* A binary tree node has data, pointer to left child and a pointer to right child */struct node{ int data; int vc; struct node *left, *right;};// A memoization based function that returns size of the minimum vertex cover.int vCover(struct node *root){ // The size of minimum vertex cover is zero if tree is empty or there // is only one node if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return 0; // If vertex cover for this node is already evaluated, then return it // to save recomputation of same subproblem again. if (root->vc != 0) return root->vc; // Calculate size of vertex cover when root is part of it int size_incl = 1 + vCover(root->left) + vCover(root->right); // Calculate size of vertex cover when root is not part of it int size_excl = 0; if (root->left) size_excl += 1 + vCover(root->left->left) + vCover(root->left->right); if (root->right) size_excl += 1 + vCover(root->right->left) + vCover(root->right->right); // Minimum of two values is vertex cover, store it before returning root->vc = min(size_incl, size_excl); return root->vc;}// A utility function to create a nodestruct node* newNode( int data ){ struct node* temp = (struct node *) malloc( sizeof(struct node) ); temp->data = data; temp->left = temp->right = NULL; temp->vc = 0; // Set the vertex cover as 0 return temp;}// Driver program to test above functionsint main(){ // Let us construct the tree given in the above diagram struct node *root = newNode(20); root->left = newNode(8); root->left->left = newNode(4); root->left->right = newNode(12); root->left->right->left = newNode(10); root->left->right->right = newNode(14); root->right = newNode(22); root->right->right = newNode(25); std::cout << "Size of the smallest vertex cover is " << vCover(root) << std::endl; return 0;} |
Java
/* Dynamic programming based program for Vertex Cover problem for a Binary Tree */class GFG{ // A utility function to find min of two integers static int min(int x, int y) { return (x < y) ? x : y; } /* * A binary tree node has data, pointer to left child and a pointer to right * child */ static class node { int data; int vc; node left, right; }; // A memoization based function that returns // size of the minimum vertex cover. static int vCover(node root) { // The size of minimum vertex cover is zero // if tree is empty or there is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // If vertex cover for this node is // already evaluated, then return it // to save recomputation of same subproblem again. if (root.vc != 0) return root.vc; // Calculate size of vertex cover // when root is part of it int size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it int size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Minimum of two values is vertex cover, // store it before returning root.vc = Math.min(size_incl, size_excl); return root.vc; } // A utility function to create a node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = temp.right = null; temp.vc = 0; // Set the vertex cover as 0 return temp; } // Driver code public static void main(String[] args) { // Let us construct tree given in the above diagram node root = newNode(20); root.left = newNode(8); root.left.left = newNode(4); root.left.right = newNode(12); root.left.right.left = newNode(10); root.left.right.right = newNode(14); root.right = newNode(22); root.right.right = newNode(25); System.out.printf("Size of the smallest vertex" + "cover is %d ", vCover(root)); }}// This code is contributed by PrinciRaj1992 |
C#
/* Dynamic programming based program for Vertex Cover problem for a Binary Tree */using System;class GFG{ // A utility function to find // min of two integers static int min(int x, int y) { return (x < y) ? x : y; } /* * A binary tree node has data, pointer to left child and a pointer to right * child */ class node { public int data; public int vc; public node left, right; }; // A memoization based function that returns // size of the minimum vertex cover. static int vCover(node root) { // The size of minimum vertex cover is zero // if tree is empty or there is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // If vertex cover for this node is // already evaluated, then return it // to save recomputation of same subproblem again. if (root.vc != 0) return root.vc; // Calculate size of vertex cover // when root is part of it int size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it int size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Minimum of two values is vertex cover, // store it before returning root.vc = Math.Min(size_incl, size_excl); return root.vc; } // A utility function to create a node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = temp.right = null; temp.vc = 0; // Set the vertex cover as 0 return temp; } // Driver code public static void Main(String[] args) { // Let us construct tree given in the above diagram node root = newNode(20); root.left = newNode(8); root.left.left = newNode(4); root.left.right = newNode(12); root.left.right.left = newNode(10); root.left.right.right = newNode(14); root.right = newNode(22); root.right.right = newNode(25); Console.Write("Size of the smallest vertex" + "cover is {0} ", vCover(root)); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>/* Dynamic programming based program forVertex Cover problem for a Binary Tree */// A utility function to find min of two integersfunction min(x,y){ return (x < y) ? x : y;}/* * A binary tree node has data, pointer to left child and a pointer to right * child */class Node{ constructor(data) { this.vc=0; this.data=data; this.left=this.right=null; }}// A memoization based function that returns// size of the minimum vertex cover.function vCover(root){ // The size of minimum vertex cover is zero // if tree is empty or there is only one node if (root == null) return 0; if (root.left == null && root.right == null) return 0; // If vertex cover for this node is // already evaluated, then return it // to save recomputation of same subproblem again. if (root.vc != 0) return root.vc; // Calculate size of vertex cover // when root is part of it let size_incl = 1 + vCover(root.left) + vCover(root.right); // Calculate size of vertex cover // when root is not part of it let size_excl = 0; if (root.left != null) size_excl += 1 + vCover(root.left.left) + vCover(root.left.right); if (root.right != null) size_excl += 1 + vCover(root.right.left) + vCover(root.right.right); // Minimum of two values is vertex cover, // store it before returning root.vc = Math.min(size_incl, size_excl); return root.vc;}// Driver code// Let us construct tree given in the above diagram let root = new Node(20); root.left = new Node(8); root.left.left = new Node(4); root.left.right = new Node(12); root.left.right.left = new Node(10); root.left.right.right = new Node(14); root.right = new Node(22); root.right.right = new Node(25); document.write("Size of the smallest vertex " + "cover is ", vCover(root));// This code is contributed by rag2127</script> |
Python3
# A binary tree node has data, pointer to left child and a pointer to# right childclass node: def __init__(self): # instance fields found by C++ to Python Converter: self.data = 0 self.vc = 0 self.left = None self.right = None# Driver program to test above functionsdef main(): # Let us construct the tree given in the above diagram root = Globals.newNode(20) root.left = Globals.newNode(8) root.left.left = Globals.newNode(4) root.left.right = Globals.newNode(12) root.left.right.left = Globals.newNode(10) root.left.right.right = Globals.newNode(14) root.right = Globals.newNode(22) root.right.right = Globals.newNode(25) print("Size of the smallest vertex cover is ", end='') print(Globals.vCover(root), end='') print("\n", end='')class Globals: # A utility function to find min of two integers @staticmethod def min(x, y): if(x < y): return x else: return y # A memoization based function that returns size of the minimum vertex cover. @staticmethod def vCover(root): # The size of minimum vertex cover is zero if tree is empty or there # is only one node if root is None: return 0 if root.left is None and root.right is None: return 0 # If vertex cover for this node is already evaluated, then return it # to save recomputation of same subproblem again. if root.vc != 0: return root.vc # Calculate size of vertex cover when root is part of it size_incl = 1 + Globals.vCover(root.left) + Globals.vCover(root.right) # Calculate size of vertex cover when root is not part of it size_excl = 0 if root.left: size_excl += 1 + \ Globals.vCover(root.left.left) + \ Globals.vCover(root.left.right) if root.right: size_excl += 1 + \ Globals.vCover(root.right.left) + \ Globals.vCover(root.right.right) # Minimum of two values is vertex cover, store it before returning root.vc = Globals.min(size_incl, size_excl) return root.vc # A utility function to create a node @staticmethod def newNode(data): temp = node() temp.data = data temp.left = temp.right = None temp.vc = 0 # Set the vertex cover as 0 return tempif __name__ == "__main__": main() |
Output
Size of the smallest vertex cover is 3
Time Complexity:
The time complexity of the vCover function is O(n), where n is the number of nodes in the binary tree. This is because each node is visited only once and its vertex cover size is calculated in constant time.
Space Complexity:
The space complexity of the program is O(n), where n is the number of nodes in the binary tree. This is because the space required for the binary tree is proportional to the number of nodes in the tree. Additionally, the memoization technique used in the vCover function requires additional space to store the vertex cover sizes of already evaluated nodes. However, since the depth of the recursion tree is limited by the height of the binary tree, the space complexity of the program is also O(h), where h is the height of the binary tree. In the worst case scenario where the binary tree is skewed, the height of the tree can be as large as n, which would result in a space complexity of O(n).
References:
http://courses.csail.mit.edu/6.006/spring11/lectures/lec21.pdf
Exercise:
Extend the above solution for n-ary trees.
This article is contributed by Udit Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Approach for any general tree :
1. Approach will be same dynamic programming approach as discussed.
2. For every node, if we exclude this node from vertex cover than we have to include its neighbouring nodes,
and if we include this node in the vertex cover than we will take the minimum of the two possibilities of taking its neighbouring
nodes in the vertex cover to get minimum vertex cover.
3. We will store the above information in the dp array.
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// An utility function to add an edge in the treevoid addEdge(vector<int> adj[], int x, int y){ adj[x].push_back(y); adj[y].push_back(x);}void dfs(vector<int> adj[], vector<int> dp[], int src, int par){ for (auto child : adj[src]) { if (child != par) dfs(adj, dp, child, src); } for (auto child : adj[src]) { if (child != par) { // not including source in the vertex cover dp[src][0] += dp[child][1]; // including source in the vertex cover dp[src][1] += min(dp[child][1], dp[child][0]); } }}// function to find minimum size of vertex covervoid minSizeVertexCover(vector<int> adj[], int N){ vector<int> dp[N + 1]; for (int i = 1; i <= N; i++) { // 0 denotes not included in vertex cover dp[i].push_back(0); // 1 denotes included in vertex cover dp[i].push_back(1); } dfs(adj, dp, 1, -1); // printing minimum size vertex cover cout << min(dp[1][0], dp[1][1]) << endl;}// Driver Codeint main(){ /* 1 / \ 2 7 / \ 3 6 / | \ 4 8 5 */ // number of nodes in the tree int N = 8; // adjacency list representation of the tree vector<int> adj[N + 1]; addEdge(adj, 1, 2); addEdge(adj, 1, 7); addEdge(adj, 2, 3); addEdge(adj, 2, 6); addEdge(adj, 3, 4); addEdge(adj, 3, 8); addEdge(adj, 3, 5); minSizeVertexCover(adj, N); return 0;} |
Java
// Java implementation for the above approachimport java.util.*;class GFG { // An utility function to add an edge in the tree static void addEdge(List<List<Integer> > adj, int x, int y) { adj.get(x).add(y); adj.get(y).add(x); } static void dfs(List<List<Integer> > adj, List<List<Integer> > dp, int src, int par) { for (Integer child : adj.get(src)) { if (child != par) dfs(adj, dp, child, src); } for (Integer child : adj.get(src)) { if (child != par) { // not including source in the vertex cover dp.get(src).set(0, dp.get(child).get(1) + dp.get(src).get(0)); // including source in the vertex cover dp.get(src).set( 1, dp.get(src).get(1) + Math.min(dp.get(child).get(1), dp.get(child).get(0))); } } } // function to find minimum size of vertex cover static void minSizeVertexCover(List<List<Integer> > adj, int N) { List<List<Integer> > dp = new ArrayList<>(); for (int i = 0; i <= N; i++) { dp.add(new ArrayList<>()); } for (int i = 1; i <= N; i++) { // 0 denotes not included in vertex cover dp.get(i).add(0); // 1 denotes included in vertex cover dp.get(i).add(1); } dfs(adj, dp, 1, -1); // printing minimum size vertex cover System.out.println( Math.min(dp.get(1).get(0), dp.get(1).get(1))); } public static void main(String[] args) { /* 1 / \ 2 7 / \ 3 6 / | \ 4 8 5 */ // number of nodes in the tree int N = 8; // adjacency list representation of the tree List<List<Integer> > adj = new ArrayList<>(); for (int i = 0; i <= N; i++) { adj.add(new ArrayList<>()); } addEdge(adj, 1, 2); addEdge(adj, 1, 7); addEdge(adj, 2, 3); addEdge(adj, 2, 6); addEdge(adj, 3, 4); addEdge(adj, 3, 8); addEdge(adj, 3, 5); minSizeVertexCover(adj, N); }} |
Python3
# Python3 implementation for the above approachdef addEdge(adj, x, y): adj[x].append(y) adj[y].append(x)def dfs(adj, dp, src, par): for child in adj[src]: if child != par: dfs(adj, dp, child, src) for child in adj[src]: if child != par: # not including source in the vertex cover dp[src][0] = dp[child][1] + dp[src][0] # including source in the vertex cover dp[src][1] = dp[src][1] + min(dp[child][1], dp[child][0])def minSizeVertexCover(adj, N): dp = [[0 for j in range(2)] for i in range(N+1)] for i in range(1, N+1): # 0 denotes not included in vertex cover dp[i][0] = 0 # 1 denotes included in vertex cover dp[i][1] = 1 dfs(adj, dp, 1, -1) # printing minimum size vertex cover print(min(dp[1][0], dp[1][1]))# Driver Code""" 1 / \ 2 7 / \ 3 6 /|\ 4 8 5"""# number of nodes in the treeN = 8# adjacency list representation of the treeadj = [[] for i in range(N+1)]addEdge(adj, 1, 2)addEdge(adj, 1, 7)addEdge(adj, 2, 3)addEdge(adj, 2, 6)addEdge(adj, 3, 4)addEdge(adj, 3, 8)addEdge(adj, 3, 5)minSizeVertexCover(adj, N) |
C#
using System;using System.Collections.Generic;class GFG{ // An utility function to add an edge in the tree static void addEdge(List<List<int>> adj, int x, int y) { adj[x].Add(y); adj[y].Add(x); } static void dfs(List<List<int>> adj, List<List<int>> dp, int src, int par) { foreach (int child in adj[src]) { if (child != par) dfs(adj, dp, child, src); } foreach (int child in adj[src]) { if (child != par) { // not including source in the vertex cover dp[src][0] = dp[child][1] + dp[src][0]; // including source in the vertex cover dp[src][1] = dp[src][1] + Math.Min(dp[child][1], dp[child][0]); } } } // function to find minimum size of vertex cover static void minSizeVertexCover(List<List<int>> adj, int N) { List<List<int>> dp = new List<List<int>>(); for (int i = 0; i <= N; i++) { dp.Add(new List<int>()); } for (int i = 1; i <= N; i++) { // 0 denotes not included in vertex cover dp[i].Add(0); // 1 denotes included in vertex cover dp[i].Add(1); } dfs(adj, dp, 1, -1); // printing minimum size vertex cover Console.WriteLine(Math.Min(dp[1][0], dp[1][1])); } public static void Main(string[] args) { /* 1 / \ 2 7 / \ 3 6 / | \ 4 8 5 */ // number of nodes in the tree int N = 8; // adjacency list representation of the tree List<List<int>> adj = new List<List<int>>(); for (int i = 0; i <= N; i++) { adj.Add(new List<int>()); } addEdge(adj, 1, 2); addEdge(adj, 1, 7); addEdge(adj, 2, 3); addEdge(adj, 2, 6); addEdge(adj, 3, 4); addEdge(adj, 3, 8); addEdge(adj, 3, 5); minSizeVertexCover(adj, N); }} |
Javascript
// Javascript implementation for the above approach // An utility function to add an edge in the tree function addEdge(adj, x, y) { adj[x].push(y); adj[y].push(x); } function dfs(adj, dp, src, par) { for (let child of adj[src]) { if (child != par) dfs(adj, dp, child, src); } for (let child of adj[src]) { if (child != par) { // not including source in the vertex cover dp[src][0] += dp[child][1]; // including source in the vertex cover dp[src][1] += Math.min(dp[child][1], dp[child][0]); } } } // function to find minimum size of vertex cover function minSizeVertexCover(adj, N) { let dp = Array.from(Array(N + 1), () => new Array()); for (var i = 1; i <= N; i++) { // 0 denotes not included in vertex cover dp[i].push(0); // 1 denotes included in vertex cover dp[i].push(1); } dfs(adj, dp, 1, -1); // printing minimum size vertex cover console.log(Math.min(dp[1][0], dp[1][1])); } // Driver Code /* 1 / \ 2 7 / \ 3 6 / | \ 4 8 5 */ // number of nodes in the tree var N = 8; // adjacency list representation of the tree let adj = Array.from(Array(N + 1), () => new Array()); addEdge(adj, 1, 2); addEdge(adj, 1, 7); addEdge(adj, 2, 3); addEdge(adj, 2, 6); addEdge(adj, 3, 4); addEdge(adj, 3, 8); addEdge(adj, 3, 5); minSizeVertexCover(adj, N); |
Output
3
Time Complexity : O(N)
Auxiliary space : O(N)
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