Maximum number of edges to be added to a tree so that it stays a Bipartite graph

A tree is always a Bipartite Graph as we can always break into two disjoint sets with alternate levels. In other words we always color it with two colors such that alternate levels have same color. The task is to compute the maximum no. of edges that can be added to the tree so that it remains Bipartite Graph.

Examples:

Input : Tree edges as vertex pairs 
        1 2
        1 3
Output : 0
Explanation :
The only edge we can add is from node 2 to 3.
But edge 2, 3 will result in odd cycle, hence 
violation of Bipartite Graph property.

Input : Tree edges as vertex pairs 
        1 2
        1 3
        2 4
        3 5
Output : 2
Explanation : On colouring the graph, {1, 4, 5} 
and {2, 3} form two different sets. Since, 1 is 
connected from both 2 and 3, we are left with 
edges 4 and 5. Since, 4 is already connected to
2 and 5 to 3, only options remain {4, 3} and 
{5, 2}.

1) Do a simple DFS (or BFS) traversal of graph and color it with two colors.
2) While coloring also keep track of counts of nodes colored with the two colors. Let the two counts be count_color0 and count_color1.
3) Now we know maximum edges a bipartite graph can have are count_color0 x count_color1.
4) We also know tree with n nodes has n-1 edges.
5) So our answer is count_color0 x count_color1 – (n-1).

Below is the implementation :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP code to print maximum edges such that
// Tree remains a Bipartite graph
#include <bits/stdc++.h>
using namespace std;
  
// To store counts of nodes with two colors
long long count_color[2];
  
void dfs(vector<int> adj[], int node, int parent, int color)
{
    // Increment count of nodes with current
    // color
    count_color[color]++;
  
    // Traversing adjacent nodes
    for (int i = 0; i < adj[node].size(); i++) {
  
        // Not recurring for the parent node
        if (adj[node][i] != parent)
            dfs(adj, adj[node][i], node, !color);
    }
}
  
// Finds maximum number of edges that can be added
// without violating Bipartite property.
int findMaxEdges(vector<int> adj[], int n)
{
    // Do a DFS to count number of nodes
    // of each color
    dfs(adj, 1, 0, 0);
  
    return count_color[0] * count_color[1] - (n - 1);
}
  
// Driver code
int main()
{
    int n = 5;
    vector<int> adj[n + 1];
    adj[1].push_back(2);
    adj[1].push_back(3);
    adj[2].push_back(4);
    adj[3].push_back(5);
    cout << findMaxEdges(adj, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java code to print maximum edges such that
// Tree remains a Bipartite graph
import java.util.*;
  
class GFG
{
  
// To store counts of nodes with two colors
static long []count_color = new long[2];
  
static void dfs(Vector<Integer> adj[], int node, 
                      int parent, boolean color)
{
    // Increment count of nodes with current
    // color
    count_color[color == false ? 0 : 1]++;
  
    // Traversing adjacent nodes
    for (int i = 0; i < adj[node].size(); i++) 
    {
  
        // Not recurring for the parent node
        if (adj[node].get(i) != parent)
            dfs(adj, adj[node].get(i), node, !color);
    }
}
  
// Finds maximum number of edges that can be added
// without violating Bipartite property.
static int findMaxEdges(Vector<Integer> adj[], int n)
{
    // Do a DFS to count number of nodes
    // of each color
    dfs(adj, 1, 0, false);
  
    return (int) (count_color[0] * 
                  count_color[1] - (n - 1));
}
  
// Driver code
public static void main(String[] args)
{
    int n = 5;
    Vector<Integer>[] adj = new Vector[n + 1];
    for (int i = 0; i < n + 1; i++) 
        adj[i] = new Vector<Integer>();
      
    adj[1].add(2);
    adj[1].add(3);
    adj[2].add(4);
    adj[3].add(5);
    System.out.println(findMaxEdges(adj, n));
}
  
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 code to print maximum edges such 
# that Tree remains a Bipartite graph 
def dfs(adj, node, parent, color):
      
    # Increment count of nodes with 
    # current color 
    count_color[color] += 1
  
    # Traversing adjacent nodes
    for i in range(len(adj[node])):
  
        # Not recurring for the parent node 
        if (adj[node][i] != parent):
            dfs(adj, adj[node][i], 
                         node, not color)
  
# Finds maximum number of edges that 
# can be added without violating 
# Bipartite property. 
def findMaxEdges(adj, n):
      
    # Do a DFS to count number of 
    # nodes of each color 
    dfs(adj, 1, 0, 0
  
    return (count_color[0] *
            count_color[1] - (n - 1))
  
# Driver code 
  
# To store counts of nodes with 
# two colors 
count_color = [0, 0
n = 5
adj = [[] for i in range(n + 1)] 
adj[1].append(2
adj[1].append(3
adj[2].append(4
adj[3].append(5
print(findMaxEdges(adj, n))
  
# This code is contributed by PranchalK

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# code to print maximum edges such that
// Tree remains a Bipartite graph
using System;
using System.Collections.Generic;
  
class GFG
{
  
// To store counts of nodes with two colors
static long []count_color = new long[2];
  
static void dfs(List<int> []adj, int node, 
                     int parent, bool color)
{
    // Increment count of nodes with current
    // color
    count_color[color == false ? 0 : 1]++;
  
    // Traversing adjacent nodes
    for (int i = 0; i < adj[node].Count; i++) 
    {
  
        // Not recurring for the parent node
        if (adj[node][i] != parent)
            dfs(adj, adj[node][i], node, !color);
    }
}
  
// Finds maximum number of edges that can be added
// without violating Bipartite property.
static int findMaxEdges(List<int> []adj, int n)
{
    // Do a DFS to count number of nodes
    // of each color
    dfs(adj, 1, 0, false);
  
    return (int) (count_color[0] * 
                  count_color[1] - (n - 1));
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 5;
    List<int> []adj = new List<int>[n + 1];
    for (int i = 0; i < n + 1; i++) 
        adj[i] = new List<int>();
      
    adj[1].Add(2);
    adj[1].Add(3);
    adj[2].Add(4);
    adj[3].Add(5);
    Console.WriteLine(findMaxEdges(adj, n));
}
}
  
// This code is contributed by PrinciRaj1992

chevron_right



Output:

2

Time Complexity: O(n)

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


2


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.