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++

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// 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

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Python3

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# 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")

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Output:

Yes


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Improved By : sanjeev2552