Number of relations that are neither Reflexive nor Irreflexive on a Set

Given a positive integer N, the task is to find the number of relations that are neither reflexive nor irreflexive on a set of first N natural numbers. Since the count of relations 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.

Examples:

Input: N = 2
Output: 8
Explanation: Considering the set {1, 2}, the total possible relations that are neither reflexive nor irreflexive are:

1. {(1, 1)}
2. {(1, 1), (1, 2)}
3. {(1, 1), (2, 1)}
4. {(1, 1), (1, 2), (2, 1)}
5. {(2, 2)}
6. {(2, 2), (2, 1)}
7. {(2, 2), (1, 2)}
8. {(2, 2), (2, 1), (1, 2)}

Therefore, the total count is 8.

Input: N = 3
Output: 384

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.
• A relation will be non-reflexive if it doesn’t contain at least one pair of (x, x) and relation will be non-irreflexive if it contains at least one pair of (x, x) where x € R.
• It can be concluded that the relation will be non-reflexive and non-irreflexive if it contains at least one pair of (x, x) and at most (N – 1) pairs of (x, x).
• Among N pairs of (x, x), the total number of possibilities of choosing any number of pairs except 0 and N – 1 is (2^N – 2). For the remaining (N² – N) elements, each element has two choices i.e., either to include or exclude it in the subset.

From the above observations, the total number of relations that are neither reflexive nor irreflexive on a set of first N natural numbers is given by .

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

const int mod = 1000000007;

// Function to calculate x^y
// modulo 10^9 + 7 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;

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

    while (y > 0) {

        // If y is odd, then
        // multiply x with res
        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 the number
// of relations that are neither
// reflexive nor irreflexive
void countRelations(int N)
{
    // Return the resultant count
    cout << (power(2, N) - 2)
                * power(2, N * N - N);
}

// Driver Code
int main()
{
    int N = 2;
    countRelations(N);

    return 0;
}

Java

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

class GFG{

static int mod = 1000000007;

// Function to calculate x^y
// modulo 10^9 + 7 in O(log y)
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 res
        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 value of x^y
    return res;
}

// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
static void countRelations(int N)
{

    // Return the resultant count
    System.out.print((power(2, N) - 2) *
                      power(2, N * N - N));
}

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

    countRelations(N);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python program for the above approach
mod = 1000000007

# Function to calculate x^y
# modulo 10^9 + 7 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

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

    while(y > 0):

        # If y is odd, then
        # multiply x with res
        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 value of x^y
    return res

# Function to count the number
# of relations that are neither
# reflexive nor irreflexive
def countRelations(N):

    # Return the resultant count
    print((power(2, N) - 2) * power(2, N * N - N))

# Driver Code
N = 2
countRelations(N)

# This code is contributed by abhinavjain194

C#

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

class GFG{

static int mod = 1000000007;

// Function to calculate x^y
// modulo 10^9 + 7 in O(log y)
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 res
        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 value of x^y
    return res;
}

// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
static void countRelations(int N)
{

    // Return the resultant count
    Console.Write((power(2, N) - 2) *
                   power(2, N * N - N));
}

// Driver Code
public static void Main(String[] args)
{
    int N = 2;
    countRelations(N);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program for the above approach

var mod = 1000000007;

// Function to calculate x^y
// modulo 10^9 + 7 in O(log y)
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 res
        if (y %2 != 0)
            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 the number
// of relations that are neither
// reflexive nor irreflexive
function countRelations(N)
{
    // Return the resultant count
    document.write((power(2, N) - 2)
                * power(2, (N * N) - N));
}

// Driver Code
var N = 2;
countRelations(N);

</script>

Output:

8

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