Find if a degree sequence can form a simple graph | Havel-Hakimi Algorithm

Given a sequence of non-negative integers arr[], the task is to check if there exists a simple graph corresponding to this degree sequence. Note that a simple graph is a graph with no self-loops and parallel edges.

Examples:

Input: arr[] = {3, 3, 3, 3}
Output: Yes
This is actually a complete graph(K_4)



Input: arr[] = {3, 2, 1, 0}
Output: No
A vertex has degree n-1 so it’s connected to all the other n-1 vertices.
But another vertex has degree 0 i.e. isolated. It’s a contradiction.

Approach: One way to check the existence of a simple graph is by Havel-Hakimi algorithm given below:

  • Sort the sequence of non-negative integers in non-increasing order.
  • Delete the first element(say V). Subtract 1 from the next V elements.
  • Repeat 1 and 2 until one of the stopping conditions is met.

Stopping conditions:

  • All the elements remaining are equal to 0 (Simple graph exists).
  • Negative number encounter after subtraction (No simple graph exists).
  • Not enough elements remaining for the subtraction step (No simple graph exists).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if
// a simple graph exists
bool graphExists(vector<int> &a, int n)
{
    // Keep performing the operations until one
    // of the stopping condition is met
    while (1)
    {
        // Sort the list in non-decreasing order
        sort(a.begin(), a.end(), greater<>());
  
        // Check if all the elements are equal to 0
        if (a[0] == 0 and a[a.size() - 1] == 0)
            return true;
  
        // Store the first element in a variable
        // and delete it from the list
        int v = a[0];
        a.erase(a.begin() + 0);
  
        // Check if enough elements
        // are present in the list
        if (v > a.size())
            return false;
  
        // Subtract first element from next v elements
        for (int i = 0; i < v; i++)
        {
            a[i]--;
  
            // Check if negative element is
            // encountered after subtraction
            if (a[i] < 0)
                return false;
        }
    }
}
  
// Driver Code
int main()
{
    vector<int> a = {3, 3, 3, 3};
    int n = a.size();
  
    graphExists(a, n) ? cout << "Yes" : 
                        cout << "NO" << endl;
  
    return 0;
}
  
// This code is contributed by
// sanjeev2552

chevron_right






                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python implementation of the approach  


  


# Function that returns true if  


# a simple graph exists 


def graphExists(a): 


  


    # Keep performing the operations until one  


    # of the stopping condition is met 


    while True:  


  


        # Sort the list in non-decreasing order 


        a = sorted(a, reverse = True) 


  


        # Check if all the elements are equal to 0 


        if a[0]== 0 and a[len(a)-1]== 0: 


            return True


  


        # Store the first element in a variable 


        # and delete it from the list 


        v = a[0] 


        a = a[1:] 


  


        # Check if enough elements  


        # are present in the list 


        if v>len(a):  


            return False


  


        # Subtract first element from next v elements 


        for i in range(v): 


            a[i]-= 1


  


            # Check if negative element is  


            # encountered after subtraction 


            if a[i]<0: 


                return False


  


  


# Driver code 


a = [3, 3, 3, 3] 


if(graphExists(a)): 


    print("Yes") 


else: 


    print("No") 



















                        chevron_right







                        filter_none











Output:


Yes






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
SujanDutta
Check out this Author's contributed articles.
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 Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  sanjeev2552