Find if a degree sequence can form a simple graph | HavelHakimi Algorithm (original) (raw)

Last Updated : 24 Jun, 2024

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 &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 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

Python

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

JavaScript

**Time Complexity: O(N^2*logN)
**Auxiliary Space: O(1)