Number of Relations that are both Irreflexive and Antisymmetric on a Set

Given a positive integer N, the task is to find the number of relations that are irreflexive antisymmetric relations that can be formed over the given set of elements.  Since the count can be very large, print it to modulo 10⁹ + 7.

A relation R on a set A is called reflexive if no (a, a) € R holds for every element a € A.
For Example: If set A = {a, b} then R = {(a, b), (b, a)} is irreflexive relation.

A relation R on a set A is called Antisymmetric if and only if (a, b) € R and (b, a) € R, then a = b is called antisymmetric, i.e., the relation R = {(a, b)? R | a ? b} is anti-symmetric, since a ? b and b ? a implies a = b.

Examples:

Input: N = 2
Output: 3
Explanation:
Considering the set {a, b}, all possible relations that are both irreflexive and antisymmetric relations are:

1. {}
2. {{a, b}}
3. {{b, a}}

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.
• No (x, x) pair should be included in the subset to make sure the relation is irreflexive.
• For the remaining (N² – N) pairs, divide them into (N² – N)/2 groups where each group consists of a pair (x, y) and its symmetric pair (y, x).
• Now, there are three choices, either include one of the ordered pairs or neither of them from a group.
• Therefore, the total number of possible relations that are both irreflexive and antisymmetric is given by 3^{(N2 – N)/2}.

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 % mod in O(log y)
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;

    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 value of x^y
    return res;
}

// Function to count relations that
// are  irreflexive and antisymmetric
// in a set consisting of N elements
int numberOfRelations(int N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

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

    return 0;
}

Java

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

class GFG{

static int mod = 1000000007;

// Function to calculate
// x ^ y % mod in O(log y)
static int power(long x, int y)
{

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

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

    while (y > 0)
    {

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

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

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

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

// Function to count relations that
// are  irreflexive and antisymmetric
// in a set consisting of N elements
static int numberOfRelations(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(numberOfRelations(N));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python 3 program for the above approach
mod = 1000000007

# Function to calculate
# x ^ y % mod in O(log y)
def power(x,  y):

    # Stores the result of x^y
    res = 1

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

    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 value of x^y
    return res

# Function to count relations that
# are  irreflexive and antisymmetric
# in a set consisting of N elements
def numberOfRelations(N):

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

# Driver Code
if __name__ == "__main__":

    N = 2
    print(numberOfRelations(N))

    # This code is contributed by ukasp.

C#

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

class GFG{

static int mod = 1000000007;

// Function to calculate
// x ^ y % mod in O(log y)
static int power(long x, int y)
{

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

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

    while (y > 0)
    {

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

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

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

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

// Function to count relations that
// are  irreflexive and antisymmetric
// in a set consisting of N elements
static int numberOfRelations(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.Write(numberOfRelations(N));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program for the above approach

let mod = 1000000007;

// Function to calculate
// x ^ y % mod in O(log y)
function power(x, y)
{
    // Stores the result of x^y
    let res = 1;

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

    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 value of x^y
    return res;
}

// Function to count relations that
// are  irreflexive and antisymmetric
// in a set consisting of N elements
function numberOfRelations(N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code

     let N = 2;
    document.write(numberOfRelations(N));

</script>

Output:

3

Time Complexity: O(log N)
Auxiliary Space: O(1),  since no extra space has been taken.