Skip to content
Related Articles

Related Articles

Find if a degree sequence can form a simple graph | Havel-Hakimi Algorithm
  • Difficulty Level : Basic
  • Last Updated : 07 Dec, 2020

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(K4)

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++




// 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)
            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

Java




// Java implementation of the approach
import java.util.*;
 
@SuppressWarnings("unchecked")
 
class GFG{
   
// Function that returns true if
// a simple graph exists
static boolean graphExists(ArrayList a, int n)
{
     
    // Keep performing the operations until one
    // of the stopping condition is met
    while (true)
    {
         
        // Sort the list in non-decreasing order
        Collections.sort(a, Collections.reverseOrder());
          
        // Check if all the elements are equal to 0
        if ((int)a.get(0) == 0)
            return true;
    
        // Store the first element in a variable
        // and delete it from the list
        int v = (int)a.get(0);
        a.remove(a.get(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.set(i, (int)a.get(i) - 1);
    
            // Check if negative element is
            // encountered after subtraction
            if ((int)a.get(i) < 0)
                return false;
        }
    }
}
   
// Driver Code
public static void main(String[] args)
{
    ArrayList a = new ArrayList();
    a.add(3);
    a.add(3);
    a.add(3);
    a.add(3);
     
    int n = a.size();
      
    if (graphExists(a, n))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("NO");
    }
}
}
 
// This code is contributed by pratham76

Python3




# Python3 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")

C#




// C# implementation of the approach
using System;
using System.Collections;
 
class GFG{
  
// Function that returns true if
// a simple graph exists
static bool graphExists(ArrayList a, int n)
{
     
    // Keep performing the operations until one
    // of the stopping condition is met
    while (true)
    {
         
        // Sort the list in non-decreasing order
        a.Sort();
        a.Reverse();
         
        // Check if all the elements are equal to 0
        if ((int)a[0] == 0)
            return true;
   
        // Store the first element in a variable
        // and delete it from the list
        int v = (int)a[0];
        a.Remove(a[0]);
   
        // Check if enough elements
        // are present in the list
        if (v > a.Count)
            return false;
   
        // Subtract first element from
        // next v elements
        for(int i = 0; i < v; i++)
        {
            a[i] = (int)a[i] - 1;
   
            // Check if negative element is
            // encountered after subtraction
            if ((int)a[i] < 0)
                return false;
        }
    }
}
  
// Driver Code
public static void Main(string[] args)
{
    ArrayList a = new ArrayList(){ 3, 3, 3, 3 };
    int n = a.Count;
     
    if (graphExists(a, n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("NO");
    }
}
}
 
// This code is contributed by rutvik_56
Output: 
Yes

 

 Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.  

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course

My Personal Notes arrow_drop_up
Recommended Articles
Page :