Number of Asymmetric Relations on a set of N elements

Given a positive integer N, the task is to find the number of Asymmetric Relations in a set of N elements.  Since the number of relations can be very large, print it modulo 10⁹+7.

A relation R on a set A is called Asymmetric if and only if x R y exists, then y R x for every (x, y) € A.
For Example: If set A = {a, b}, then R = {(a, b)} is asymmetric relation.

Examples:

Input: N = 2
Output: 3
Explanation: Considering the set {1, 2}, the total number of possible asymmetric relations are {{}, {(1, 2)}, {(2, 1)}}.

Input: N = 5
Output: 59049

Approach: The given problem can be solved based on the following observations:

• A relation R on a set A is a subset of the Cartesian product of a set, i.e. A * A with N² elements.
• There are total N pairs of type (x, x) that are present in the Cartesian product, where any of (x, x) should not be included in the subset.
• Now, one is left with (N² – N) elements of the Cartesian product.
• To satisfy the property of asymmetric relation, one has three possibilities of either to include only of type (x, y) or only of type (y, x) or none from a single group into the subset.
• Hence, the total number of possible asymmetric relations is equal to 3 ^{(N2 – N) / 2}.

Therefore, the idea is to print the value of 3^{(N2 – N)/2} modulo 10⁹ + 7 as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

const int mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
int power(long long x,
          unsigned int y)
{
    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0) {

        // If y is odd, then
        // multiply x with result
        if (y & 1)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
int asymmetricRelation(int N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
int main()
{
    int N = 2;
    cout << asymmetricRelation(N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

final static int mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
public static int power(int x, int y)
{

    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0)
    {

        // If y is odd, then
        // multiply x with result
        if (y % 2 == 1)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
public static int asymmetricRelation(int N)
{

    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver code
public static void main (String[] args)
{
    int N = 2;

    System.out.print(asymmetricRelation(N));
}
}

// This code is contributed by user_qa7r

Python3

# Python3 program for the above approach
mod = 1000000007

# Function to calculate
# x^y modulo (10^9 + 7)
def power(x, y):

    # Stores the result of x^y
    res = 1

    # Update x if it exceeds mod
    x = x % mod

    # If x is divisible by mod
    if (x == 0):
        return 0

    while (y > 0):

        # If y is odd, then
        # multiply x with result
        if (y & 1):
            res = (res * x) % mod;

        # Divide y by 2
        y = y >> 1

        # Update the value of x
        x = (x * x) % mod

    # Return the final
    # value of x ^ y
    return res

# Function to count the number of
# asymmetric relations in a set
# consisting of N elements
def asymmetricRelation(N):

    # Return the resultant count
    return power(3, (N * N - N) // 2)

# Driver Code
if __name__ == '__main__':

    N = 2

    print(asymmetricRelation(N))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;

class GFG{

const int mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
static int power(int x, int y)
{

    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0)
    {

        // If y is odd, then
        // multiply x with result
        if ((y & 1) != 0)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
static int asymmetricRelation(int N)
{

    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
public static void Main(string[] args)
{
    int N = 2;
    Console.WriteLine(asymmetricRelation(N));
}
}

// This code is contributed by ukasp

Javascript

<script>

// Javascript program for the above approach

var mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
function power(x, y)
{
    // Stores the result of x^y
    var res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0) {

        // If y is odd, then
        // multiply x with result
        if (y & 1)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
function asymmetricRelation( N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
var N = 2;
document.write( asymmetricRelation(N));


</script>

Output:

3

Time Complexity: O(N*log N)
Auxiliary Space: O(1)