Number of Simple Graph with N Vertices and M Edges (original) (raw)

Last Updated : 15 Sep, 2022

Given two integers N and M, the task is to count the number of simple undirected graphs that can be drawn with N vertices and M edges. A simple graph is a graph that does not contain multiple edges and self-loops.

Examples:

Input: N = 3, M = 1
Output: 3
The 3 graphs are {1-2, 3}, {2-3, 1}, {1-3, 2}.

Input: N = 5, M = 1
Output: 10

Approach: The N vertices are numbered from 1 to N. As there are no self-loops or multiple edges, the edge must be present between two different vertices. So the number of ways we can choose two different vertices is NC2 which is equal to (N * (N - 1)) / 2. Assume it P.

Now M edges must be used with these pairs of vertices, so the number of ways to choose M pairs of vertices between P pairs will be PCM.

If P < M then the answer will be 0 as the extra edges can not be left alone.

Below is the implementation of the above approach:

C++ `

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;

// Function to return the value of // Binomial Coefficient C(n, k) int binomialCoeff(int n, int k) {

if (k > n)
    return 0;

int res = 1;

// Since C(n, k) = C(n, n-k)
if (k > n - k)
    k = n - k;

// Calculate the value of
// [n * (n - 1) *---* (n - k + 1)] / [k * (k - 1) * ... * 1]
for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
}

return res;

}

// Driver Code int main() { int N = 5, M = 1;

int P = (N * (N - 1)) / 2;

cout << binomialCoeff(P, M);

return 0;

}

Java

// Java implementation of the approach class GFG {

// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{

    if (k > n)
        return 0;

    int res = 1;

    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;

    // Calculate the value of
    // [n * (n - 1) *---* (n - k + 1)] /
    // [k * (k - 1) * ... * 1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}

// Driver Code public static void main(String[] args) { int N = 5, M = 1; int P = (N * (N - 1)) / 2;

System.out.println(binomialCoeff(P, M));

} }

// This code is contributed by Shivi_Aggarwal

Python 3

Python 3 implementation of the approach

Function to return the value of

Binomial Coefficient C(n, k)

def binomialCoeff(n, k):

if (k > n):
    return 0

res = 1

# Since C(n, k) = C(n, n-k)
if (k > n - k):
    k = n - k

# Calculate the value of
# [n * (n - 1) *---* (n - k + 1)] / 
# [k * (k - 1) * ... * 1]
for i in range( k):
    res *= (n - i)
    res //= (i + 1)

return res

Driver Code

if name=="main":

N = 5
M = 1

P = (N * (N - 1)) // 2

print(binomialCoeff(P, M))

This code is contributed by ita_c

C#

// C# implementation of the approach using System;

class GFG { // Function to return the value of // Binomial Coefficient C(n, k) static int binomialCoeff(int n, int k) {

if (k > n)
    return 0;

int res = 1;

// Since C(n, k) = C(n, n-k)
if (k > n - k)
    k = n - k;

// Calculate the value of
// [n * (n - 1) *---* (n - k + 1)] / 
// [k * (k - 1) * ... * 1]
for (int i = 0; i < k; ++i)
{
    res *= (n - i);
    res /= (i + 1);
}

return res;

}

// Driver Code public static void Main() { int N = 5, M = 1;

int P = (N * (N - 1)) / 2;

Console.Write(binomialCoeff(P, M));

} }

// This code is contributed // by Akanksha Rai

PHP

$n) return 0; $res = 1; // Since C(n, k) = C(n, n-k) if ($k > n−n - n−k) k=k = k=n - $k; // Calculate the value of // [n * (n - 1) *---* (n - k + 1)] / // [k * (k - 1) * ... * 1] for ($i = 0; i<i < i<k; ++$i) { res∗=(res *= (res∗=(n - $i); res/=(res /= (res/=(i + 1); } return $res; } // Driver Code $N = 5; $M = 1; P=floor((P = floor((P=floor((N * ($N - 1)) / 2); echo binomialCoeff($P, $M); // This code is contributed by Ryuga ?>

JavaScript

`

Complexity Analysis: