Counting Graphs

To determine the total number of edges possible in a graph with 'N' vertices, we can use the combination formula. Each edge requires two vertices so we can choose any two vertices from the 'N' vertices. Therefore, the total number of edges possible, denoted as 'E', is given by the formula E = (N * (N-1)) / 2.

Now, when counting the total number of graphs, we can consider that each edge in the graph may either exist or not exist. This gives us two options for each edge. Since there are 'E' edges, the total number of possible combinations of edge choices is 2^E. This represents all the different possible graphs that can be formed.

// To find the total number of "undirected" graphs by N vertices

// can be acheived by 2^(N(N-1)/2);

int countingGraphs(int N)

{

    // Write your code here.

    int power=N*(N-1)/2;

    return pow(2,power);

}

from typing import *

def countingGraphs(N: int) -> int:

    # Write your code here

    if N==0 or N==1:

        return N

    return  round(2**(N * (N - 1) / 2))

    return  

    pass

int countingGraphs(int N)

{

    int p=N*(N-1)/2;

    return pow(2,p);// Write your code here.

}

public class Solution {

public static int countingGraphs(int N) {

if (N == 0 || N == 1) {

return N;

}

return (int) Math.round(Math.pow(2, N * (N - 1) / 2));

}

}

int factorial(int n){

  int fact=1;

  while (n > 0) {

    fact = fact * n;

    n = n - 1;

  }

}

int countingGraphs(int N)

{

    if (N==1){

        return 1;

    }

    if(N==2){

        return 2;

    }

    return pow(2,factorial(N)/(2*factorial(N-2)));

}

Number of Graphs = 2^(n*(n-1)/2)

int countingGraphs(int N){
    return pow(2,N*(N-1)/2);
}

int powval = N*(N-1)/2;

        return (int)Math.pow(2, powval);

public class Solution {

    public static int countingGraphs(int N) {

        double pow=N*(N-1)/2;

        double cnt=Math.pow(2, pow);

        return (int)cnt;

    }

}

number of edges =nC2

number of graph = 2^(nC2)

int countingGraphs(int N)
{
   int nC2=N*(N-1)/2;
   return 1<<(nC2);
}