Count of distinct graphs that can be formed with N vertices

Given an integer N which is the number of vertices. The task is to find the number of distinct graphs that can be formed. Since the answer can be very large, print the answer % 1000000007.

Examples:

Input: N = 3
Output: 8



Input: N = 4
Output: 64

Approach:

  • The maximum number of edges a graph with N vertices can contain is X = N * (N – 1) / 2.
  • The total number of graphs containing 0 edge and N vertices will be XC0
  • The total number of graphs containing 1 edge and N vertices will be XC1
  • And so on from number of edges 1 to X with N vertices
  • Hence, the total number of graphs that can be formed with n vertices will be:
    XC0 + XC1 + XC2 + … + XCX = 2X.

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;
  
const int MOD = 1e9 + 7;
  
// Function to return (x^y) % MOD
// in O(log(y))
long long power(long long x,
                long long y,
                const int& MOD)
{
    long long res = 1;
    while (y > 0) {
        if (y & 1)
            res = (res * x) % MOD;
        x = (x * x) % MOD;
        y /= 2;
    }
    return res;
}
  
// Function to return the count of distinct
// graphs possible with n vertices
long long countGraphs(int n)
{
  
    // Maximum number of edges for a
    // graph with n vertices
    long long x = n * (n - 1) / 2;
  
    // Function to calculate
    // (2^x) % mod
    return power(2, x, MOD);
}
  
// Driver code
int main()
{
    int n = 5;
  
    cout << countGraphs(n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG 
{
    static final int MOD = (int)1e9 + 7
      
    // Function to return (x^y) % MOD 
    // in O(log(y)) 
    static long power(long x, 
                      long y)
    
        long res = 1
        while (y > 0
        
            if ((y & 1) != 0
                res = (res * x) % MOD; 
            x = (x * x) % MOD; 
            y /= 2
        
        return res; 
    
      
    // Function to return the count of distinct 
    // graphs possible with n vertices 
    static long countGraphs(int n) 
    
      
        // Maximum number of edges for a 
        // graph with n vertices 
        long x = n * (n - 1) / 2
      
        // Function to calculate 
        // (2^x) % mod 
        return power(2, x); 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int n = 5
      
        System.out.println(countGraphs(n)); 
    }
}
  
// This code is contributed by AnkitRai01

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Python

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MOD = int(1e9 + 7)
  
# Function to return the count of distinct
# graphs possible with n vertices
def countGraphs(n):
  
    # Maximum number of edges for a
    # graph with n vertices
    x = (n *( n - 1 )) //2
      
    # Return 2 ^ x
    return (pow(2, x, MOD))
  
# Driver code
n = 5
print(countGraphs(n))

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C#

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// C# implementation of the approach 
using System;
  
class GFG
{
    static int MOD = (int)1e9 + 7; 
      
    // Function to return (x^y) % MOD 
    // in O(log(y)) 
    static long power(long x, long y)
    
        long res = 1; 
        while (y > 0) 
        
            if ((y & 1) != 0) 
                res = (res * x) % MOD; 
            x = (x * x) % MOD; 
            y /= 2; 
        
        return res; 
    
      
    // Function to return the count of distinct 
    // graphs possible with n vertices 
    static long countGraphs(int n) 
    
      
        // Maximum number of edges for a 
        // graph with n vertices 
        long x = n * (n - 1) / 2; 
      
        // Function to calculate 
        // (2^x) % mod 
        return power(2, x); 
    
      
    // Driver code 
    static public void Main ()
    {
        int n = 5; 
      
        Console.Write(countGraphs(n)); 
    }
}
  
// This code is contributed by ajit.

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Output:

1024


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Improved By : AnkitRai01, jit_t