Find two disjoint good sets of vertices in a given graph

Given an undirected unweighted graph with N vertices and M edges. The task is to find two disjoint good sets of vertices. A set X is called good if for every edge UV in the graph at least one of the endpoint belongs to X(i.e, U or V or both U and V belongs to X).
If it is not possible to make such sets then print -1.

Examples:

Input :

Output : {1 3 4 5} ,{2 6}
One disjoint good set contains vertices {1, 3, 4, 5} and other contains {2, 6}.

Input :

Output : -1

Approach:
One of the observation is that there is no edge UV that U and V are in the same set.The two good sets form a bipartition of the graph, so the graph has to be bipartite. And being bipartite is also sufficient. Read about bipartition here.

Below is the implementation of the above approach :

C++

// C++ program to find two disjoint 
// good sets of vertices in a given graph 
#include <bits/stdc++.h> 
using namespace std; 
#define N 100005 
  
// For the graph 
vector<int> gr[N], dis[2]; 
bool vis[N]; 
int colour[N]; 
bool bip; 
  
// Function to add edge 
void Add_edge(int x, int y) 
{ 
    gr[x].push_back(y); 
    gr[y].push_back(x); 
} 
  
// Bipartie function 
void dfs(int x, int col) 
{ 
    vis[x] = true; 
    colour[x] = col; 
  
    // Check for child vertices 
    for (auto i : gr[x]) { 
  
        // If it is not visited 
        if (!vis[i]) 
            dfs(i, col ^ 1); 
  
        // If it is already visited 
        else if (colour[i] == col) 
            bip = false; 
    } 
} 
  
// Function to find two disjoint 
// good sets of vertices in a given graph 
void goodsets(int n) 
{ 
    // Initially assume that graph is bipartie 
    bip = true; 
  
    // For every unvisited vertex call dfs 
    for (int i = 1; i <= n; i++) 
        if (!vis[i]) 
            dfs(i, 0); 
  
    // If graph is not bipartie 
    if (!bip) 
        cout << -1; 
    else { 
  
        // Differentiate two sets 
        for (int i = 1; i <= n; i++) 
            dis[colour[i]].push_back(i); 
  
        // Print vertices belongs to both sets 
  
        for (int i = 0; i < 2; i++) { 
  
            for (int j = 0; j < dis[i].size(); j++) 
                cout << dis[i][j] << " "; 
            cout << endl; 
        } 
    } 
} 
  
// Driver code 
int main() 
{ 
    int n = 6, m = 4; 
  
    // Add edges 
    Add_edge(1, 2); 
    Add_edge(2, 3); 
    Add_edge(2, 4); 
    Add_edge(5, 6); 
  
    // Function call 
    goodsets(n); 
} 

Java

// Java program to find two disjoint  
// good sets of vertices in a given graph  
import java.util.*; 
  
class GFG  
{ 
  
    static int N = 100005; 
  
    // For the graph 
    @SuppressWarnings("unchecked") 
    static Vector<Integer>[] gr = new Vector[N],  
                            dis = new Vector[2]; 
    static
    { 
        for (int i = 0; i < N; i++) 
            gr[i] = new Vector<>(); 
        for (int i = 0; i < 2; i++) 
            dis[i] = new Vector<>(); 
    } 
    static boolean[] vis = new boolean[N]; 
    static int[] color = new int[N]; 
    static boolean bip; 
  
    // Function to add edge 
    static void add_edge(int x, int y) 
    { 
        gr[x].add(y); 
        gr[y].add(x); 
    } 
  
    // Bipartie function 
    static void dfs(int x, int col)  
    { 
        vis[x] = true; 
        color[x] = col; 
  
        // Check for child vertices 
        for (int i : gr[x]) 
        { 
  
            // If it is not visited 
            if (!vis[i]) 
                dfs(i, col ^ 1); 
  
            // If it is already visited 
            else if (color[i] == col) 
                bip = false; 
        } 
    } 
  
    // Function to find two disjoint 
    // good sets of vertices in a given graph 
    static void goodsets(int n) 
    { 
        // Initially assume that graph is bipartie 
        bip = true; 
  
        // For every unvisited vertex call dfs 
        for (int i = 1; i <= n; i++) 
            if (!vis[i]) 
                dfs(i, 0); 
  
        // If graph is not bipartie 
        if (!bip) 
            System.out.println(-1); 
        else
        { 
  
            // Differentiate two sets 
            for (int i = 1; i <= n; i++) 
                dis[color[i]].add(i); 
  
            // Print vertices belongs to both sets 
  
            for (int i = 0; i < 2; i++) 
            { 
                for (int j = 0; j < dis[i].size(); j++) 
                    System.out.print(dis[i].elementAt(j) + " "); 
                System.out.println(); 
            } 
        } 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        int n = 6, m = 4; 
  
        // Add edges 
        add_edge(1, 2); 
        add_edge(2, 3); 
        add_edge(2, 4); 
        add_edge(5, 6); 
  
        // Function call 
        goodsets(n); 
    } 
} 
  
// This code is contributed by 
// sanjeev2552 

Python 3

# Python 3 program to find two disjoint 
# good sets of vertices in a given graph 
N = 100005
  
# For the graph 
gr = [[] for i in range(N)] 
dis = [[] for i in range(2)] 
vis = [False for i in range(N)] 
colour = [0 for i in range(N)] 
bip = 0
  
# Function to add edge 
def Add_edge(x, y): 
    gr[x].append(y) 
    gr[y].append(x) 
  
# Bipartie function 
def dfs(x, col): 
    vis[x] = True
    colour[x] = col 
  
    # Check for child vertices 
    for i in gr[x]: 
          
        # If it is not visited 
        if (vis[i] == False): 
            dfs(i, col ^ 1) 
  
        # If it is already visited 
        elif (colour[i] == col): 
            bip = False
  
# Function to find two disjoint 
# good sets of vertices in a given graph 
def goodsets(n): 
      
    # Initially assume that  
    # graph is bipartie 
    bip = True
  
    # For every unvisited vertex call dfs 
    for i in range(1, n + 1, 1): 
        if (vis[i] == False): 
            dfs(i, 0) 
  
    # If graph is not bipartie 
    if (bip == 0): 
        print(-1) 
    else: 
          
        # Differentiate two sets 
        for i in range(1, n + 1, 1): 
            dis[colour[i]].append(i) 
  
        # Print vertices belongs to both sets 
        for i in range(2): 
            for j in range(len(dis[i])): 
                print(dis[i][j], end = " ")  
            print('\n', end = "") 
  
# Driver code 
if __name__ == '__main__': 
    n = 6
    m = 4
  
    # Add edges 
    Add_edge(1, 2) 
    Add_edge(2, 3) 
    Add_edge(2, 4) 
    Add_edge(5, 6) 
  
    # Function call 
    goodsets(n) 
  
# This code is contributed 
# by Surendra_Gangwar 

Output:

1 3 4 5 
2 6

My Personal Notes

C++

Java

Python 3

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